Could a Gyroscopic inertial thruster ever work?

[Red Iron Crown]

If you're going to give someone grief about mixing up radius and diameter, you should probably stop using the word "orbit" to describe what's happening in the gyroscope experiment. It is not gravity that is causing the system to rotate, so it's not an orbit at all.

[Dodgey]

You have claimed that you have done the expirment yourself, why not film it and show us the results of the video.

[Momentus]

If you're going to give someone grief about mixing up radius and diameter, you should probably stop using the word "orbit" to describe what's happening in the gyroscope experiment. It is not gravity that is causing the system to rotate, so it's not an orbit at all.

Whilst I see your point about the use of the word orbit, I am not trying to win a debate, but to explain. circular movement is clumsy don't you think? As you have clearly understood what was meant, I think I will continue to use orbit in this context.

A minor point. without gravity, the system would not rotate.

I felt that I was being given grief over my typo, unless of course you are addressing Z Man. Do you have an opinion on momentum being a critical factor in calculating deflection?

You have claimed that you have done the expirment yourself, why not film it and show us the results of the video.

How sad is that. You need a video of a weight on a string? use a mirror.

Edited March 30, 2014 by Vanamonde

[Z-Man]

Great deconstruction of the experiment, which you could not be assed to actualy try. Pity because had you done so you could have corrected the blatant typo (Rad not Diam Doh). We would then have the true and attested behaviour of a mass orbiting a fixed point. This would have included the observation that you, being a seasoned experimenter, ensured that L and ̉ۡ did not vary.

Oh, so I misinterpreted your vague instructions there. The more important points still stand. And now, by the way, ̉ۡ varies. With a fixed string length, it increases as the movement radius increases, it has no upper bound. No way around it.

"only valid under these conditions: You have a gyro precessing" does that mean it is not an orbiting mass?

It is a mass going in circles, yes, but not in itself a gyroscope. The formula specifically requires a gyroscope that is spinning all by itself, suspended on a string attached to it off-center. The weight is not itself spinning around its center of mass, so the formula does not apply.

In the Laithwaite ice tower, did a gyro behave differently from an inert mass?

Not a gyro on a string. And of course it did behave differently. An inert mass, forced onto the same relative motion to the tower, would just have toppled over.

My issue with the K^2 is not one of mathematics but of logic. If the momentum of the mass is unknown, the radius of the orbit cannot be calculated. That is true for any mass be it spinning or inert.

The radius can be obtained for the very specific case that the pendulum is in a purely precession driven state. The state you instructed us to consider exclusively ("Then do not push it!"). The magnitude of the linear momentum at any given time in that state is fixed given the fixed specified variables. Edited March 29, 2014 by Z-Man

[Red Iron Crown]

Do you have an opinion on momentum being a critical factor in calculating deflection?

No, I've never formally studied gyroscopic motion so I'm reluctant to offer an opinion. I am learning about it from some of the posters in this thread, though, which is why I'm still following it.

[KerikBalm]

Am I the only one whom is reminded of arguing with creationists?

One side brings many facts and sound logic, the other side blatently misrepresents the facts and uses bad logic.

He's still talking about a video where you can clearly see the tower revolving about a point between it and the gyroscope, as expected.

The only thing that changing the relative masses of the tower and gyroscope will do, is move the point that the tower revolves about - either closer to the tower, or closer to the gryoscope.

Thus light vs heavy tower would only change the radius of the revolutions of the tower.

The video used masses such that the radius is easy to discern - it doesn't much matter if the radius is 1cm, or 10 cm, both can be clearly seen (but if it was 10mm, it would be too hard to see).

Many labs weren't even interested in debunking the "arsenic life" paper, because the conclusion was on such a shoddy foundation.

I'm surprised a lab at cambridge (note that its not as if the entire university oversaw this) would even bother to put any effort to debunk this. Why... they might as well be trying to debunk some quack claim of alchemy.

Its Russel's teapot all over again.

The burden of proof is on you Momentus.

Showing us a video of a tower of unknown mass revolving around a point that is quite plausibly the CG does not support your claim.

Referring to an experiment of a tower with legs quite capable of digging into the ice it is sitting on, does not support your claim.

We have no reason to believe your claims, and no burden to disprove them.

[Dodgey]

How sad is that. You need a video of a weight on a string? use a mirror.

No, I need proof of a gyroscope precessing around a point on a free moving tower that is not its center of mass.

[K^2]

My issue with the K^2 is not one of mathematics but of logic. If the momentum of the mass is unknown, the radius of the orbit cannot be calculated.

That's because you don't understand any physics. I actually know the amount of momentum that gyro will have. It follows from equations of motion.

So why are YOU refusing to do the actual experiment described?

You need a video of a weight on a string? use a mirror.

No, I need video of gyro on a string. I KNOW how a pendulum works. YOU don't know how a gyro works. We need you to try the actual experiment described, because you obviously have no idea what you are talking about.

Edited March 29, 2014 by K^2

[M Drive]

I'm getting ahead of myself here, but I have a somewhat interesting question to you educated like folks.

Say my machine passes the pendulum test, and assume we eventually find out the propulsion is the same effect seen in Alex Jones' machine. Given those clues, how would you try to fit this phenomenon into our own physics? What would be the most likely explanation that we've somehow missed all this time?

[Dodgey]

Well, firstly the expirment would be recreated in a much more controlled environment, say in a vacuume, and the machine would be remade with much more precision and documentation and as simply as possible to determine the actual forces at work here. After that I have no clue, mass celebration in the streets maybe? Bring alcohole.

[K^2]

Yeah. If it passes pendulum test for you, I am out of obvious ideas. I would have to replicate it and start excluding variables with more experiments and using sensors to try and figure out where all the forces are.

[Momentus]

Am I the only one whom is reminded of arguing with creationists?

One side brings many facts and sound logic, the other side blatently misrepresents the facts and uses bad logic.

He's still talking about a video where you can clearly see the tower revolving about a point between it and the gyroscope, as expected.

The only thing that changing the relative masses of the tower and gyroscope will do, is move the point that the tower revolves about - either closer to the tower, or closer to the gryoscope.

Thus light vs heavy tower would only change the radius of the revolutions of the tower.

The video used masses such that the radius is easy to discern - it doesn't much matter if the radius is 1cm, or 10 cm, both can be clearly seen (but if it was 10mm, it would be too hard to see).

Many labs weren't even interested in debunking the "arsenic life" paper, because the conclusion was on such a shoddy foundation.

I'm surprised a lab at cambridge (note that its not as if the entire university oversaw this) would even bother to put any effort to debunk this. Why... they might as well be trying to debunk some quack claim of alchemy.

Its Russel's teapot all over again.

The burden of proof is on you Momentus.

Showing us a video of a tower of unknown mass revolving around a point that is quite plausibly the CG does not support your claim.

Referring to an experiment of a tower with legs quite capable of digging into the ice it is sitting on, does not support your claim.

We have no reason to believe your claims, and no burden to disprove them.

I have no interest in your belief structure. You can accept the doctrine you have been taught by those whom you think are older and wiser, agree with the view of the masses, name and shame heretics.

The experiment does not care what you believe, neither do I feel any obligation to teach you. There is no one on the site who cares enough about what you say to correct the more obvious errors, why should I.

Momentus

Oh, so I misinterpreted your vague instructions there. The more important points still stand. And now, by the way, ̉ۡ varies. With a fixed string length, it increases as the movement radius increases, it has no upper bound. No way around it.

It is a mass going in circles, yes, but not in itself a gyroscope. The formula specifically requires a gyroscope that is spinning all by itself, suspended on a string attached to it off-center. The weight is not itself spinning around its center of mass, so the formula does not apply.

Not a gyro on a string. And of course it did behave differently. An inert mass, forced onto the same relative motion to the tower, would just have toppled over.

The radius can be obtained for the very specific case that the pendulum is in a purely precession driven state. The state you instructed us to consider exclusively ("Then do not push it!"). The magnitude of the linear momentum at any given time in that state is fixed given the fixed specified variables.

You did not do the simple string experiment, although experiment is rather a grand name for it. The statement that " ̉ۡ varies. With a fixed string length, it increases as the movement radius increases, it has no upper bound."Does not come from your actual experience. It is a canard of magnificent proportions. Perhaps no one reads 'what you wrote'.

Momentus

I'm getting ahead of myself here, but I have a somewhat interesting question to you educated like folks.

Say my machine passes the pendulum test, and assume we eventually find out the propulsion is the same effect seen in Alex Jones' machine. Given those clues, how would you try to fit this phenomenon into our own physics? What would be the most likely explanation that we've somehow missed all this time?

I hesitate to taint you by replying to your post. If you read what has been said up to now, there will be no attempt to understand what you have done, only a mad chorus of "experimental error" at best and "charlatan" at least. That there is a flaw in the application of Newton's laws on conservation is the explanation.

Momentus

No, I need proof of a gyroscope precessing around a point on a free moving tower that is not its center of mass.

Then you should look at video 5.

Edited March 30, 2014 by Vanamonde

[Dodgey]

There is not enough data for us to determine with any sort of accuracy where the center of mass is in that set up. Thus we can not make the assumption that it is not rotating around the center of mass.

[Momentus]

That's because you don't understand any physics. I actually know the amount of momentum that gyro will have. It follows from equations of motion.

So why are YOU refusing to do the actual experiment described?

No, I need video of gyro on a string. I KNOW how a pendulum works. YOU don't know how a gyro works. We need you to try the actual experiment described, because you obviously have no idea what you are talking about.

If you did know how a pendulum works, then you would know that the size of the orbit radius is a function of the linear momentum initially imparted to the mass. Without that initial momentum the mass will hang directly below the point of suspension, not moving at all. If you had reached that point then the next question would have been where does that linear momentum come from and where does it appear in your formula.

And unless and until some other member can take up the burden of explaining the physics of a pendulum? I did a number of experiments with gyros 20 years ago. If you really want to know about the anomaly, then buy a good gyro and invest some time and effort in your own education. Stop thinking that you can learn science by scoring debating points. The experiment is the only arbiter of truth.

Momentus.

Momentus

There is not enough data for us to determine with any sort of accuracy where the center of mass is in that set up. Thus we can not make the assumption that it is not rotating around the center of mass.

I can show you the available evidence, I cannot make you think about it. If you think this video was faked, then more of the same is not going to penetrate your deep knowledge of the subject

Edited March 30, 2014 by Vanamonde

[Dodgey]

Have you determined the location of the center of mass in that setup. Without actual figures that video is useless.

EDIT: A better question is how have you determined the location of the center of the mass.

Edited March 30, 2014 by Dodgey

[K^2]

If you did know how a pendulum works, then you would know that the size of the orbit radius is a function of the linear momentum initially imparted to the mass. Without that initial momentum the mass will hang directly below the point of suspension, not moving at all. If you had reached that point then the next question would have been where does that linear momentum come from and where does it appear in your formula.

And unless and until some other member can take up the burden of explaining the physics of a pendulum? I did a number of experiments with gyros 20 years ago. If you really want to know about the anomaly, then buy a good gyro and invest some time and effort in your own education. Stop thinking that you can learn science by scoring debating points. The experiment is the only arbiter of truth.

Momentus.

Momentus

Yet you refuse to do the experiment. Hm.

And for the tenth time, there are two separate oscillations here. One due to pendulum and one due to gyro. Not giving the system a "push", id est, having zero pendulum-caused oscillation, was one of your own requirements.

For the sake of more educated here, the problem is governed by a second order differential equation. Working with complex numbers to represent 2D displacement of center of mass, call it r, and the point where string supports the gyro, call it z, we can define:

r = z + de^ỉۡt

Here, d is distance between the point on axis where we attach the string of length L to the center of mass of gyro with mass m. Since string is the only source of force, equation of motion is very simple.

mr'' = -mzg/L

Note that this equation completely ignores torque on the gyro that may be caused by the string. This is a valid assumption under some conditions I'm going to discuss later. The most general solution, one that includes gyro-pendulum interactions, is extremely complex, chaotic, and is similar to a double-pendulum problem. Fortunately, we don't have to deal with all that.

Masses cancel, and we substitute z = r - de^ỉۡt.

r'' + rg/L = de^ỉۡt g/L

At this point, it should be obvious that this is a driven harmonic oscillator. As with any inhomogeneous linear equation, we can start by solving the homogeneous part.

r₀'' + r₀ Ω² = 0

Here, I substituted Ω² = g/L for simplicity, which leads to a general solution.

r₀ = ae^iΩt + be^-iΩt

For some pair of complex numbers a and b to be determined from boundary condition or other information. So now we need a particular solution for the inhomogeneous part. Naturally, it has to have e^iÉt term, so we go with a trial solution r₁ = ce^iÉt. Substituting that into differential equation, we have an algebraic equation.

cΩ² - cɲ = dΩ²

c = dΩ²/(Ω²-ɲ)

And since a general solution is just sum of general homogeneous solution and particular solution, r₀+r₁, we have solution to the system.

r = ae^iΩt + be^-iΩt + de^iÉtΩ²/(Ω²-ɲ)

This gives us center of mass location as a function of time. It's clear that there are two separate oscillations going on there. One has frequency Ω and is identical to the motion of pendulum. Relative to it, there is a second oscillation with frequency É, which is frequency of gyro's precession.

So the only question left is the values for a and b. In general, they can be anything. All you have to do is give the right amount of push to the system from the start. But Momentus wanted to talk about the case where gyro precesses around an almost vertical string. That is the case with minimal energy of the system. Just for sake of completeness, let us write down the energy of the system. It contains a kinetic term and a potential energy term.

E = mr'²/2 + z²g/(2L)

If we consider r = r₀ + r₁, since r₁ has no parameters, by triangle inequality, the above is minimized when the above equation is minimized for r₀. And since it is trivially 0 for a = b = 0 and cannot be negative, that is our minimum.

In other words, the absolute minimum amount the pendulum-gyro system can have is solution with a = b = 0

r = de^iÉtΩ²/(Ω²-ɲ)

Which happens to be identical to our particular solution. In this case, we can look at motion of the suspension point, rather than center of mass. Substitution the above into equation for z, we have:

z = de^iÉtΩ²/(Ω²-ɲ) - de^iÉt = de^iÉtɲ/(Ω²-ɲ)

Naturally, if you just want absolute displacement, we are looking at the norm of z.

|z| = dɲ/(Ω²-ɲ) = dLɲ/(g - ɲL)

And, as I have pointed out earlier, this also satisfies the condition on centrifugal force.

The above equation is clearly singular when Ω²=ɲ. In fact, there are serious problems when two are even similar. ɲ << Ω² results in positive displacement, which just counters centrifugal force as gyro slowly precesses. However, the case where ɲ >> Ω² is also entirely valid. Here, center of mass stays relatively put near the center, as the suspension point rapidly spins around. That's a case of a weak gyro, one which Momentus doesn't seem to even consider. But it is also a valid solution.

One should be able to guess that when ɲ ~ Ω², the z and r are not colinear. That is a serious problem. As I've mentioned near the start, we are ignoring torque on the gyro. That's perfectly fine so long as z and r have the same direction. There really is no torque. Gyro stays horizontal and continues to precess at a constant rate. But if orientation of z and r do not match, which is going to be the case for near-resonance or for a bad choice of parameters a and b, this assumption falls apart. worst part is that this torque will cause gyro to turn towards vertical. As that happens, center of mass lowers or rises, changing potential energy stored in gyro. As a result, you have energy exchange between pendulum and gyro. As I've mentioned, it's similar to interaction in a double-pendulum and is chaotic.

So general solution, for general starting conditions or general relation between É and Ω, is not available. But provided for a far-from-resonance condition and initial condition with minimal energy ("no push"), we do have an exact solution which does give us a fixed displacement provided here and in earlier posts.

The most important part to take from all of this is that it is impossible to have oscillations with two different "orbits". This is not a simple pendulum. There is one solution with a = b = 0. And any significant deviation from that, because É and Ω do not match, result in a significant angle between z and r, causing gyro to do weird things.

So again, there is only one solution of the type Momentus wants to talk about, giving unique displacement, which he'd be able to confirm if he actually bothered to do the experiment rather than be an annoying troll.

Edited March 30, 2014 by K^2

[Momentus]

Yet you refuse to do the experiment. Hm.

And for the tenth time, there are two separate oscillations here. One due to pendulum and one due to gyro. Not giving the system a "push", id est, having zero pendulum-caused oscillation, was one of your own requirements.

For the sake of more educated here, the problem is governed by a second order differential equation. Working with complex numbers to represent 2D displacement of center of mass, call it r, and the point where string supports the gyro, call it z, we can define:

r = z + de^ỉۡt

Here, d is distance between the point on axis where we attach the string of length L to the center of mass of gyro with mass m. Since string is the only source of force, equation of motion is very simple.

mr'' = -mzg/L

Note that this equation completely ignores torque on the gyro that may be caused by the string. This is a valid assumption under some conditions I'm going to discuss later. The most general solution, one that includes gyro-pendulum interactions, is extremely complex, chaotic, and is similar to a double-pendulum problem. Fortunately, we don't have to deal with all that.

Masses cancel, and we substitute z = r - de^ỉۡt.

r'' + rg/L = de^ỉۡt g/L

At this point, it should be obvious that this is a driven harmonic oscillator. As with any inhomogeneous linear equation, we can start by solving the homogeneous part.

r₀'' + r₀ Ω² = 0

Here, I substituted Ω² = g/L for simplicity, which leads to a general solution.

r₀ = ae^iΩt + be^-iΩt

For some pair of complex numbers a and b to be determined from boundary condition or other information. So now we need a particular solution for the inhomogeneous part. Naturally, it has to have e^iÉt term, so we go with a trial solution r₁ = ce^iÉt. Substituting that into differential equation, we have an algebraic equation.

cΩ² - cɲ = dΩ²

c = dΩ²/(Ω²-ɲ)

And since a general solution is just sum of general homogeneous solution and particular solution, r₀+r₁, we have solution to the system.

r = ae^iΩt + be^-iΩt + de^iÉtΩ²/(Ω²-ɲ)

This gives us center of mass location as a function of time. It's clear that there are two separate oscillations going on there. One has frequency Ω and is identical to the motion of pendulum. Relative to it, there is a second oscillation with frequency É, which is frequency of gyro's precession.

So the only question left is the values for a and b. In general, they can be anything. All you have to do is give the right amount of push to the system from the start. But Momentus wanted to talk about the case where gyro precesses around an almost vertical string. That is the case with minimal energy of the system. Just for sake of completeness, let us write down the energy of the system. It contains a kinetic term and a potential energy term.

E = mr'²/2 + z²g/(2L)

If we consider r = r₀ + r₁, since r₁ has no parameters, by triangle inequality, the above is minimized when the above equation is minimized for r₀. And since it is trivially 0 for a = b = 0 and cannot be negative, that is our minimum.

In other words, the absolute minimum amount the pendulum-gyro system can have is solution with a = b = 0

r = de^iÉtΩ²/(Ω²-ɲ)

Which happens to be identical to our particular solution. In this case, we can look at motion of the suspension point, rather than center of mass. Substitution the above into equation for z, we have:

z = de^iÉtΩ²/(Ω²-ɲ) - de^iÉt = de^iÉtɲ/(Ω²-ɲ)

Naturally, if you just want absolute displacement, we are looking at the norm of z.

|z| = dɲ/(Ω²-ɲ) = dLɲ/(g - ɲL)

And, as I have pointed out earlier, this also satisfies the condition on centrifugal force.

The above equation is clearly singular when Ω²=ɲ. In fact, there are serious problems when two are even similar. ɲ << Ω² results in positive displacement, which just counters centrifugal force as gyro slowly precesses. However, the case where ɲ >> Ω² is also entirely valid. Here, center of mass stays relatively put near the center, as the suspension point rapidly spins around. That's a case of a weak gyro, one which Momentus doesn't seem to even consider. But it is also a valid solution.

One should be able to guess that when ɲ ~ Ω², the z and r are not colinear. That is a serious problem. As I've mentioned near the start, we are ignoring torque on the gyro. That's perfectly fine so long as z and r have the same direction. There really is no torque. Gyro stays horizontal and continues to precess at a constant rate. But if orientation of z and r do not match, which is going to be the case for near-resonance or for a bad choice of parameters a and b, this assumption falls apart. worst part is that this torque will cause gyro to turn towards vertical. As that happens, center of mass lowers or rises, changing potential energy stored in gyro. As a result, you have energy exchange between pendulum and gyro. As I've mentioned, it's similar to interaction in a double-pendulum and is chaotic.

So general solution, for general starting conditions or general relation between É and Ω, is not available. But provided for a far-from-resonance condition and initial condition with minimal energy ("no push"), we do have an exact solution which does give us a fixed displacement provided here and in earlier posts.

The most important part to take from all of this is that it is impossible to have oscillations with two different "orbits". This is not a simple pendulum. There is one solution with a = b = 0. And any significant deviation from that, because É and Ω do not match, result in a significant angle between z and r, causing gyro to do weird things.

So again, there is only one solution of the type Momentus wants to talk about, giving unique displacement, which he'd be able to confirm if he actually bothered to do the experiment rather than be an annoying troll.

That is an excellent analysis. Your conclusion that the Rotation of the gyro about its own centre of mass which is precession whilst moving with orbital motion of a pendulum means that in any situation, the behaviour must be chaotic - with the exception being when as you say Ω²=ɲ. - this gives the locked behaviour as when the moon orbits the earth. This is a complex interaction, difficult to explain, you did a good job.

But and there has to be one I'm sorry to say this not the way a gyro behaves. Indeed one of the fascinations is its smooth motion in a true circle.

I actually went to the toy store and got a moderately suitable gyro and tried that. When suspended on a string, it behaves as calculated within the wide limits of observational accuracy. Its slowest precession is about one revolution in two seconds, that's slow enough to be slower than the oscillation given the 40-50 cm rope length I used. Unfortunately, the bearings are crap and it drops down to resonant precession within about ten seconds,

swinging damps out then moves smoothly. Troll the internet, Google gyro pictures, there are no examples of the chaotic behaviour that should be there. No sarcasm. Your analysis is spot on, the behaviour should be chaotic. But it is not. I was trying to make that point when I said that taking a gyro that is in a stable orbit and shorten the string should result in change.

My insistence that the gyro is not pushed, is to ensure that no external force is applied to set the gyro in orbit. Under those conditions, the gyro should remain in the same position rotating (precessing) about its own C of G until some external force is transmitted along the string. For pure precession there is no conflict between the two modes because the C of G cannot move.

Yet again this is not how a gyro behaves. No matter how carefully it is released the offset supported gyro will not simply precess about the C of G.

Your careful and I hasten to say correct analysis of gyro behaviour is not the real world behaviour. I doubt if any other physicist has taken the trouble to do this.

Momentus

[Z-Man]

Reread the analysis. Behaviour is supposed to get complicated near the resonance. Far off the resonance either way, behaviour is expected to be more regular. Which is precisely what I observed (and, of course, others before me) and reported.

You did not do the simple string experiment, although experiment is rather a grand name for it. The statement that " ̉ۡ varies. With a fixed string length, it increases as the movement radius increases, it has no upper bound."Does not come from your actual experience. It is a canard of magnificent proportions. Perhaps no one reads 'what you wrote'.

I've swung around enough weights on strings in my life to know how they behave. But fine, if you insist to waste precious bits on the internet on an irrelevant side point of a side point, I've done it. Rope length was 60 cm, weight was suspended 3 cm off its center of mass, mass was 84g in case it matters to you. Swung it around. Counted how many rounds it did over the course of a minute.

Radius of about 25 cm: rotation frequency was 39.5 per minute (̉ۡ = 2.07/s)

Radius of somewhere between 40 and 50 cm: rotation frequency was 46 per minute (̉ۡ = 2.41/s)

Radius of somewhere close to 55 cm: rotation frequency was 60 per minute (̉ۡ = 3.14/s)

That was about as fast as I dared to fling the thing around, the rope was from the other experiment and thus not chosen for durability or painless handling. Anyway, clearly not constant.

[Momentus]

Reread the analysis. Behaviour is supposed to get complicated near the resonance. Far off the resonance either way, behaviour is expected to be more regular. Which is precisely what I observed (and, of course, others before me) and reported.

I've swung around enough weights on strings in my life to know how they behave. But fine, if you insist to waste precious bits on the internet on an irrelevant side point of a side point, I've done it. Rope length was 60 cm, weight was suspended 3 cm off its center of mass, mass was 84g in case it matters to you. Swung it around. Counted how many rounds it did over the course of a minute.

Radius of about 25 cm: rotation frequency was 39.5 per minute (̉ۡ = 2.07/s)

Radius of somewhere between 40 and 50 cm: rotation frequency was 46 per minute (̉ۡ = 2.41/s)

Radius of somewhere close to 55 cm: rotation frequency was 60 per minute (̉ۡ = 3.14/s)

That was about as fast as I dared to fling the thing around, the rope was from the other experiment and thus not chosen for durability or painless handling. Anyway, clearly not constant.

Oh dear. I talk about Pendulums, you talk about swing ball. Is there no length you wont go to just to make your debating point??

[shynung]

Oh dear. I talk about Pendulums, you talk about swing ball. Is there no length you wont go to just to make your debating point??

Are swing balls and pendulums any different?

[Z-Man]

I merely followed your instructions. You did not specify an upper limit to the force to put into the pendulum, and the example values you gave for r vs L were already well outside the limits of the linear approximation. Does not matter. Your experiment was irrelevant anyway.

K^2, about the potentially chaotic behaviour for arbitrary initial conditions in the case of É << Ω: Don't worry about it as long as d << L. Do an energy and angular momentum (z-axis, naturally) conservation analysis. If the gyro tilts up or down, it transfers both energy and angular momentum into the pendulum degrees of freedom. And if it is fast spinning, the pendulum degrees of freedom can't take the large amount of angular momentum they would need to pick up while being constrained by the amount of energy they get. It really works out beautifully and precisely: If you ignore d and start with a horizontal gyro, it is locked into the horizontal plane by the conservation laws precisely if É < Ω. Edit: for finite values of d/L, I suppose you still get narrow enough limits on the gyro's stilt to be able to ignore it.

Edit: I realized that the strict result only applies for initial states with zero energy in the pendulum degrees of motion. No matter. For É << Ω, the conservation laws still provide narrow boundaries on the allowed inclination of the gyro.

Edited March 30, 2014 by Z-Man

[Vanamonde]

Guys, if you think of something else you want to say right after posting, please just edit the post and add to it rather than submit multiple posts in a row. Sequential posting just forces others to click through more page-loads to read the same content.

[K^2]

K^2, about the potentially chaotic behaviour for arbitrary initial conditions in the case of É << Ω: Don't worry about it as long as d << L..

Yes, I was thinking that pendulum oscillations should just precess with gyro precession frequency. But I do not know how to prove it. I will run a full simulation to see what really happens in these cases. Just to make sure.

And yes, as usual, Momentus got it completely backwards. I give up on him.

[Doozler]

I give up on him.

You've been going all weekend so I think that's allowed . Your dedication is impressive!

[Momentus]

Yes, I was thinking that pendulum oscillations should just precess with gyro precession frequency. But I do not know how to prove it. I will run a full simulation to see what really happens in these cases. Just to make sure.

And yes, as usual, Momentus got it completely backwards. I give up on him.

My Hypotheses is that Newtonian Dynamics does not describe behaviour of the offset Gyroscope, as demonstrated by video 5. This has been dismissed as being due to a misunderstanding of the purpose of the experiment resulting in the heavy object being exchanged for the light one. Having done the experiment myself I know that this is not so.

The Gyroscope pendulum is another example where the analysis which is correct by the current paradigm does not predict the actual outcome.

To start right at the beginning with a spinning gyro suspended from two strings. One string is burnt through leaving the gyro suspended, point of suspension directly above the point of attachment, distance from there to the Centre of mass d. The gyro couple transfers the reaction to support the weight and the gyro remains horizontal.

The gyro precesses, which means that it turns about its own centre of gravity. It cannot displace itself. There must be an external force applied in direction of the displacement to accelerate the mass from rest. As the gyro is attached externally by the string, the displacing force must be provided by the string.

For the string to provide a horizontal component of force, it must deviate from vertical. This should be visible, initially the gyro stays still then gradually accelerates as the subtended angle increases.

It is only after the string has provided this acceleration that the gyro can move in an orbit.

There is a way to demonstrate this conclusively. At some point the ratio of d to L will be such that the horizontal force will be too small to displace the mass and the gyro will remain in position with the string circling around it. So, if L then is reduced, would the point of rotation move away from the Gyro?. Would it be a gradual process? Can L be fixed at a length such that the point of rotation would be at d/2? When does K^2's prediction become valid and z become determinate?

All of those questions are theoretical. The problem is that it does not happen that way, there is no significant deviation from vertical and the gyro mass moves immediately, it is not accelerated by the string.

Some thoughts rather than abuse would be different. Maybe make an effort to understand rather than chant the mantra, Momentus is stupid to think that known science could just possibly be wrong.

Momentus .