Could a Gyroscopic inertial thruster ever work?

[Technical Ben]

No, you can't impart momentum on an object without something pushing against the object other than itself, or having the object push something other than itself. The object will wobble back and forth, but it won't accelerate.

Just to be clear, propellant-less propulsion isn't a problem, such as solar sails, magnetic sails, gravitational slingshots, and possibly quantum thrusters. Reactionless thrusters in turn, aren't physically possible. You can't turn a rotary motion into forward motion without expelling mass either.

You could spin the device so strongly it disintegrates, then push that rubbish out of an exhaust port... a bit like a sander/grinding wheel. But that's "cheating" as it's using true fuel.

[Red Iron Crown]

I think that the encouragement offered to the M drive is cynical in the extreme. His video will not convince you that Non Newtonian motion can exist. It simply gives you the opportunity to criticise the experimental techniques.

Honestly, I suspect that better experimental techniques are all that stands to be learned from M-Drive's experiments, which isn't necessarily a bad thing. I will always encourage someone to use scientific methods to test things, even if I think those things won't work. And there's always that infinitesimal chance that he'll discover something new.

M-Drive did his own experiment and posted his results. We gave feedback on why the results may not be what he suspected they were, and offered suggestions on how to improve the experiment, even though most of us think the experiment will show nothing anomalous. There's nothing cynical about encouraging scientific rigor.

[Momentus]

Oh, yes. Here it comes again. This is a conspiracy by the scientific community! Duh!

Again, you keep making up stories of possible conspiracy, but you haven't provided any evidence that a machine violating conservation of momentum exists. I'm not saying violating Newtonian Physics, because I personally know plenty of devices that do that. Ever used a GPS? Yeah. That thing takes into account the fact that time on a GPS satellite flows at a different rate than on Earth. Due to both the gravitational field of Earth and relative motion of satellite and receiver. It simply wouldn't work without that correction. Total violation of Newtonian Physics. Except, conservation of momentum is a central result in General Relativity as well. Moreso, momentum is part of the stress-energy tensor, which is a conserved charge of Poincare local symmetry. This isn't even General Relativity, we are talking about general field theory.

I understand that these are all consepts way, way over your head. You are still stuck on freshman mechanics. But you seriously need to start getting through your head that you are going after the concept which is the most fundamental principle in all of physics. Conservation of momentum isn't some observational law, like Newton's Laws. It follows from mathematical theorems based on most fundamental symmetries of space-time structure. Other predictions of these theories are tested to 12 decimal places, both from measurements in GR and Quantum Mechannics. It's something that we know holds true for elementary particles and neutron stars in other gallaxies as a matter of fact.

And you're trying to prove it wrong with third-hand account of a bad experiment with a gyro? I feel bad for you. In all honesty, and with no offense meant. I've given you equations to work with, and if you feel you need to prove it to yourself, you should have everything you need. But if you are still under an illusion that you understand something that some of us don't, you are simply wrong, and nothing you've brought up so far has shown anything other than your ignorance of the subject.

Conspiracy theories came from some other posting. My view on suchlike is that Government is collectively too stupid to manage that. Because the experiment does not contain exotic concepts (which I can spell even if I do not understand) is it way way under your head? Is it in fact too simple for a trained scientific mind to encompass? Heavy weight swings round light support, no problem. Put some dark matter on the light bit and the heavy thing becomes the light thing, exactly the same experiment, job done. There you are, now do you understand? Or is changing the light bit to a heavy bit using imaginary stuff not scientific?

You do have one thing right in your pretentious drivel. Conservation of momentum is the basis for two of the three great symmetries. This subject is more than you can begin to comprehend.

[K^2]

Is it in fact too simple for a trained scientific mind to encompass? Heavy weight swings round light support, no problem.

Again. You have not demonstrated that. You only allude to 3rd party having claimed to have demonstrated it in dubious experiments.

I am not claiming that heavy gyro going around light tower would not be proof. I'm saying that this did not happen. Experiment you are describing never took place. Video you have posted is of a heavy tower and light gyro. And the experiment on ice was not performed under frictionless conditions. It's that simple.

I'll see if I can get the air track setup at uni to try and run heavy-gyro-light-base experiment for you. I think, it'd look even more impressive with tower being able to move only in 1D, dancing back and forward under the gyro that stays still. The problem is, ironically, finding a suitable gyro.

Conservation of momentum is the basis for two of the three great symmetries.

Cue the new age crap. You'll start talking about crystals next, right?

[Red Iron Crown]

Momentus, who exactly can understand it? Apparently I'm too dense to get it, but K^2 is too smart to comprehend it.

[Momentus]

Did not. I told you. The suspended gyroscope still acts like a regular pendulum if you push it. It differs from the regular pendulum by the two additional relevant degrees of freedom driven mainly by precession (there is interaction between them and the regular pendulum degrees of freedom). Those degrees of freedom are already there in the regular pendulum (take one, rotate the weight about the x axis, release), of course, it's just that they are strongly damped and thus not normally relevant.

There are practical reasons. On an air table, a light tower would easily get rotated sideways, causing uneven air flow and unwanted thrust. On the flip side, there is no reason to consider the Cambridge experiment irrelevant. If the claim is that a light tower is not moved by a precessing gyro, surely a heavy tower would budge even less.

Did So. still acts like a regular pendulum if you push it. Then do not push it. Then you have a mass orbiting about a point directly below the point of suspension. It is held in orbit by centripetal force imparted by the supporting string. Shorten the string and that force increases tightening the orbit. TO CONSERVE MOMENTUM the rotational speed (RPM) must increase. It does not. Ok. Ok. don't want to spoil your fun, you can push it if you must. Same deal, it still orbits at precession speed not pendulum period.

If the claim is that a light tower is not moved by a precessing gyro, surely a heavy tower would budge even less Wow, you are even better at this than K^2. Orwellian in scope. If you are going to lie, lie the big one."There are No Tanks in Bagdad". So in order to prove the claim lets make the tower heavier!! stroke of genius. Heh what about nailing the tower to the bench, then we wont have any of this rubbish about relative masses.

[K^2]

Momentus, gyro suspended from a string will turn around point ROUGHLY bellow point of suspension, UNLESS gyro precession frequency resonates with pendulum frequency. It's a trivial thing to derive. This is a driven oscillator problem, and driving force out of resonance will not push pendulum far from equilibrium point. This is, seriously, an Undergraduate level mechanics problem.

The reason this does not violate conservation of momentum is because small deflections will occur. Point of rotation will not be PERFECTLY under suspension point, resulting in small angle of the string, which is going to provide horizontal component of the force. Again, trivially derived.

Edit:

In fact, given a gyroscope with precession frequency ̉ۡ suspended from the string of length L making distance d from suspension point to center of mass, the deflection of the suspension point from center is going to be given by:

z = d Lɲ/(g - Lɲ)

Clearly, if you take a very slowly precessing gyro, with ̉ۡ small, the deflection z will also be very small. But it is not zero.

P.S. Almost forgot. This is taken in the limit of É being not very large. Specifically, ɲ << g/L. In other words, precession frequency much lower than pendulum frequency. This is to avoid resonance. A more general expression can be derived, but I'm bored with this problem already.

Edited March 28, 2014 by K^2

[Momentus]

I'll see if I can get the air track setup at uni to try and run heavy-gyro-light-base experiment for you. I think, it'd look even more impressive with tower being able to move only in 1D, dancing back and forward under the gyro that stays still. The problem is, ironically, finding a suitable gyro.

Running it on a rail is a great idea. You will be able to spin the gyro around the tower by hand and get a direct comparison with the precession motion. Comparing the spin/no spin that way will level the effects of friction. Take care if you approach the Uni. authorities with this. The reaction may well be more extreme than you expect.

http://www.gyroscope.com is a good site for gyroscopes. Needless to say This experiment will change your life. May want to give that some thought too. None the less I look forward to finally seeing an honest replication of Laithwaite experiment.

Cue the new age crap. You'll start talking about crystals next, right?

The physical symmetries of time, place, rotation. Applies to all matter, not just crystals:confused:

Momentus

[Z-Man]

Then you have a mass orbiting about a point directly below the point of suspension. It is held in orbit by centripetal force imparted by the supporting string.

Show that it happens precisely so. I say it won't. You'd observe a complicated overlay motion instead, one component looking a bit like you describe, the other being the regular pendulum motion with low amplitude. In the initial conditions you describe (gyro axis horizontal, string exactly vertical) that amplitude may (I have not done a full analysis, but edit: as K^2 said, the regular pendulum motion can be considered driven by the precession) get lower as the rotation speed of the gyro increases, so if you ever did the experiment yourself (holding the string in your hand, I'd assume), you probably did not see it.

And of course the precession speed is independent of the string length. Why would it not be?

Heh what about nailing the tower to the bench, then we wont have any of this rubbish about relative masses.

If the bench still moves in that case, your claim would be equally refuted. Point is, the heavy tower moves. Do you understand how F=ma works?

Edited March 28, 2014 by Z-Man

[K^2]

The physical symmetries of time, place, rotation. Applies to all matter, not just crystals:confused:

You forgot boosts. Oh, and I've replied to your gyro pendulum "conundrum" above with an actual equation. Get back to me when you manage to verify it.

[frizzank]

Its already possible in ksp and the real world to have this device. Use the ASAS put two arms stretched out to the side with pods and decouplers attached. Now spin up to a high velocity, then detach. You now have accelerated yourself from a closed system. Much like swinging a rock on a string and letting it go. Which incidental would also be possible in space.

The downside is your essentially throwing something from your spacecraft to gain velocity which isn't new for space travel.

Edited March 28, 2014 by frizzank

[Momentus]

You forgot boosts. Oh, and I've replied to your gyro pendulum "conundrum" above with an actual equation. Get back to me when you manage to verify it.

Point the first. "point ROUGHLY bellow point of suspension" you may wish to rephrase this. The moving pendulum bob will travel in a regular ellipse, centred EXACTLY below the point of suspension.

The bob wants to move in a straight line. The string pulls it constantly to the centre, over a full rotation, all the forces on the point of suspension must be symmetrical, ie equal and opposite. So Again " Point of rotation will not be PERFECTLY under suspension point, resulting in small angle of the string," this should be: "Centre of mass will not be PERFECTLY under suspension point, resulting in small angle of the string," More "O" level than undergrad.

And of course none of the above even now that I have corrected it has any bearing on C of M (Conservation of Momentum)

You have formulated deflection using L and ̉ۡ. Newtonian dynamics says that cannot be done. Simple logic of a pendulum is that unless the length of the pendulum is changed, the period of oscillation is a constant. No exceptions. Immutable. Fixed.

When you try to speed up a pendulum, then all that happens is the deflection increases. Try to slow it down and the radius reduces. The deflection radius therefore depends upon the momentum of the mass (Mv), not solely upon L or ̉ۡ. As v is a function (̉ۡ z ) your formulae needs more boring modification.

This is taken in the limit of angle from vertical being not very large.

For any pendulum that is in a steady state of rotation, be it Plumb bob or gyro, L ̉ۡ M and v must remain constant to maintain this steady state. If therefore a change is made to L, the length, then there will be a change in the one or more of the other values to a maintain a new state of equilibrium. As Mv must be conserved, then ̉ۡ must be the value that changes.

More if you need clarification,

Momentus

Edited March 28, 2014 by Momentus
signature

[shynung]

Its already possible in ksp and the real world to have this device. Use the ASAS put two arms stretched out to the side with pods and decouplers attached. Now spin up to a high velocity, then detach. You now have accelerated yourself from a closed system. Much like swinging a rock on a string and letting it go. Which incidental would also be possible in space.

The downside is your essentially throwing something from your spacecraft to gain velocity which isn't new for space travel.

I think that's a sling, not a gyroscope. In that case, reaction wheels rotate the whole spacecraft, and the spacecraft jettisons a part of it to launch the other elsewhere. That would not be a closed system, since it expels the rotating arms, effectively using it as reaction mass.

Edited March 28, 2014 by shynung

[KerikBalm]

Nah, we don't need clarification, its very clear you don't understand what you are talking about.

[Momentus]

Nah, we don't need clarification, its very clear you don't understand what you are talking about.

That you for your contribution. Clearly expressed, but wrong.

[K^2]

Point the first. "point ROUGHLY bellow point of suspension" you may wish to rephrase this. The moving pendulum bob will travel in a regular ellipse, centred EXACTLY below the point of suspension.

The bob wants to move in a straight line. The string pulls it constantly to the centre, over a full rotation, all the forces on the point of suspension must be symmetrical, ie equal and opposite. So Again " Point of rotation will not be PERFECTLY under suspension point, resulting in small angle of the string," this should be: "Centre of mass will not be PERFECTLY under suspension point, resulting in small angle of the string," More "O" level than undergrad.

And of course none of the above even now that I have corrected it has any bearing on C of M (Conservation of Momentum)

You have formulated deflection using L and ̉ۡ. Newtonian dynamics says that cannot be done. Simple logic of a pendulum is that unless the length of the pendulum is changed, the period of oscillation is a constant. No exceptions. Immutable. Fixed.

When you try to speed up a pendulum, then all that happens is the deflection increases. Try to slow it down and the radius reduces. The deflection radius therefore depends upon the momentum of the mass (Mv), not solely upon L or ̉ۡ. As v is a function (̉ۡ z ) your formulae needs more boring modification.

This is taken in the limit of angle from vertical being not very large.

For any pendulum that is in a steady state of rotation, be it Plumb bob or gyro, L ̉ۡ M and v must remain constant to maintain this steady state. If therefore a change is made to L, the length, then there will be a change in the one or more of the other values to a maintain a new state of equilibrium. As Mv must be conserved, then ̉ۡ must be the value that changes.

More if you need clarification,

Momentus

And you have completely missed the point in your own experiment.

̉ۡ and Sqrt(g/L) are two distinct frequencies. ̉ۡ is frequency of precession, which has NOTHING to do with the frequency of pendulum oscillation. In fact, ̉ۡ = mgd/L [This would be the angular momentum L, not length of string L. Pardon the confusion.], exactly as I've described before. Again, d is the distance between center of mass and the point where the string is attached to the gyro axis.

And yes, a pendulum moves in an ellipse about the equilibrium point. But we aren't interested in that. We are interested in motion of a gyro suspended on a string. Ground state, no push, or whatever you want to call this condition. Just the precession of the gyro. And guess what, there is still deflection. The point where string is attached to the axis of the gyro is going to be shifted from center by distance z that I have specified.

The expression can actually be simplified further, and I suggest you use that for an experiment, if we note that Lɲ in denominator is small. So if we take natural frequency of pendulum to be Ω, so that Ω² = g/L, then the displacement is:

z = d ɲ/Ω²

Does that make it cleaner for you? I suggest you go and test this. Take a string of sufficient length so that É/Ω isn't too small, and actually measure the displacement. Gyro WILL orbit point directly bellow the point of suspension with frequency É, and the point where string is attached to axis WILL be displaced from vertical by z.

Edited March 28, 2014 by K^2

[Alchemist]

I'll see if I can get the air track setup at uni to try and run heavy-gyro-light-base experiment for you. I think, it'd look even more impressive with tower being able to move only in 1D, dancing back and forward under the gyro that stays still. The problem is, ironically, finding a suitable gyro.

So, the gyro will be mostly oscillating along a line perpendicular to the track, but that means the horizontal component of the reaction force won't be always parallel to horizontal projection of the axis that means the pecession will change its tilt... I'd say the highest point will be furthest from the rail, so that the movement of the base will be of larger amplitude that the perpendicular to rail movement of the gyro. Wait, energy conservation says the same - trade between kinetic energy of gyroscope's translational movement across the rail versus it's potential energy in gravity field. Still looking at this should be mush more interesting than simple 2d circular movement

Edited March 28, 2014 by Alchemist

[Momentus]

And you have completely missed the point in your own experiment.

Absolutely I am labouring under the impression that any mass, spinning or not, moving in a circle is subjected to centripetal acceleration. There must be a force acting on that mass, which is external to the mass. With a pendulum, be it spinning or not the force is transmitted by the string, and is proportional to the angle subtended by the string from vertical. Pause for reality check

Momentus

Edited March 28, 2014 by Momentus
Missing bracket

[Z-Man]

Yes. That is precisely the basis for what K^2 calculated.

Sorry, this may be a language barrier thing. Every one of Momentus' statements falls into one of three categories:

1. Yeah, that's what mainstream physics says.

2. No, that's absolutely not how this works.

3. Wait, what is he trying to say?

So, Momentus: Please, what exactly are you saying?

z = d Lɲ/(g - Lɲ)

Actually, the only assumption you need for this to hold away from the actual resonance is that L is large compared to d and z. It will correctly describe the semi-steady state where z and d are parallel. Of course, the full dynamics get really messy near the resonance and it requires fine tuning to actually get the system into that state, but hey.

And as usual, ̉ۡ needs to be large compared to the spin frequency of the gyro.

[Momentus]

Yes. That is precisely the basis for what K^2 calculated.

Excellent. Ok so assuming that the gyro is orbiting at precession speed at radius of z . Happily turning round and round. The horizontal force component which holds it at this radius is determined by the subtended angle formed by the length of the string and the radius of displacement.

Now for a difficult manoeuvre. Increase the speed of the mass and at the same time, move it out to a larger radius orbit, such that ̉ۡ remains the same. This increases the subtended angle, which in turn increases the horizontal force. But the increase in z means that more force is required to hold the mass in orbit. Guess what, do the math and the increase in horizontal force matches the increase required to hold the new orbit. Isn't science wonderful.

z = d Lɲ/(g - Lɲ)??

Momentus

[Z-Man]

move it out to a larger radius orbit, such that ̉ۡ remains the same.

You mean with ̉ۡ as the precession angular velocity? That is impossible in general. If you try and release the gyro, you'll just have it in a mixed state of regular pendulum motion and precession driven motion.

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, so its movement is nowhere near stable enough to measure the displacement. But the qualitative behavior is as expected, sign change of z included.

z = d Lɲ/(g - Lɲ)??

K^2's formula for the displacement z of the suspension point, where d is the horizontal distance of the CoM from the suspension point, L the rope length, g Earth's surface gravity, ̉ۡ as above. Derived for a semi-steady state where, when seen from above, the gyro rotates around the center (the vertical axis through the rope's upper suspension point. That is an Ansatz. Try calculating it yourself. It gives one mode of movement (edit: technically, it's a special solution to an inhomogeneous differential equation here, I'm calling it a mode because it's actually variable if you allow different inclinations of the gyro), the other is the regular pendulum motion. Edited March 28, 2014 by Z-Man

[K^2]

Excellent. Ok so assuming that the gyro is orbiting at precession speed at radius of z . Happily turning round and round. The horizontal force component which holds it at this radius is determined by the subtended angle formed by the length of the string and the radius of displacement.

Now for a difficult manoeuvre. Increase the speed of the mass and at the same time, move it out to a larger radius orbit, such that ̉ۡ remains the same. This increases the subtended angle, which in turn increases the horizontal force. But the increase in z means that more force is required to hold the mass in orbit. Guess what, do the math and the increase in horizontal force matches the increase required to hold the new orbit. Isn't science wonderful.

z = d Lɲ/(g - Lɲ)??

Momentus

Center of mass is at r = z+d. So centrifugal force is m(z+d)ɲ. The force due to displacement is -mgz/L. I'll let you do the math. Or you can solve differential equation for motion of the gyroscope, which is what I did, and get this as a particular solution for when the whole thing isn't oscillating. General solution includes precession with this displacement about the normal oscillations of the pendulum. Something you really need to know how to derive before you argue with people on such topics.

Once again, have fun doing the math. Or better yet, do the actual experiment and take measurements. The equation I've given is absolutely correct.

[Momentus]

Center of mass is at r = z+d. So centrifugal force is m(z+d)ɲ. The force due to displacement is -mgz/L. I'll let you do the math. Or you can solve differential equation for motion of the gyroscope, which is what I did, and get this as a particular solution for when the whole thing isn't oscillating. General solution includes precession with this displacement about the normal oscillations of the pendulum. Something you really need to know how to derive before you argue with people on such topics.

Once again, have fun doing the math. Or better yet, do the actual experiment and take measurements. The equation I've given is absolutely correct.

assuming the equation you have given is z = d Lɲ/(g - Lɲ)

Ok the experiment. Do try this at home children. Take a 60 cm Length of string, with a 4oz weight tied on the end(old set of scales). Hold in right hand and swung around, at É rad/sec. gestimate diameter of circle at 20 cm. Increase agitation of right hand, gestimate new radius at 25cm. stop and repeat until boredom sets in or you run out of radii to estimate. During the experiment it will be noted that L, É and M do not vary. yet z does. Which value changes in the right hand site of your equation z = d Lɲ/(g - Lɲ) to reflect the experimental values for z?

Yes. That is precisely the basis for what K^2 calculated.

Sorry, this may be a language barrier thing. Every one of Momentus' statements falls into one of three categories:

1. Yeah, that's what mainstream physics says.

2. No, that's absolutely not how this works.

3. Wait, what is he trying to say?

So, Momentus: Please, what exactly are you saying?

This experiment is Item 1

K^2's formulae is item 2

I am unable to proceed to item 3 until there is some agreement on item 1

Momentus

[Z-Man]

assuming the equation you have given is z = d Lɲ/(g - Lɲ)

Ok the experiment. Do try this at home children. Take a 60 cm Length of string, with a 4oz weight tied on the end(old set of scales). Hold in right hand and swung around, at É rad/sec. gestimate diameter of circle at 20 cm. Increase agitation of right hand, gestimate new radius at 25cm. stop and repeat until boredom sets in or you run out of radii to estimate. During the experiment it will be noted that L, É and M do not vary. yet z does. Which value changes in the right hand site of your equation z = d Lɲ/(g - Lɲ) to reflect the experimental values for z?

The equation was specifically only valid under these conditions:

You have a gyro precessing with angular velocity ̉ۡ. There is no such thing in your setup.

z is small compared to L. Not the case in your experiment, they are about equal.

The formula is for the movement component induced by the precession. In your experiment, you are using the other, regular pendulum motion mode.

So yeah, of course the formula would describe the experiment incorrectly. The experiment does not even come close to testing the formula.

Furthermore, contrary to your claims, L (the RELEVANT rope length.Total rope length does not matter, what does matter is the length from the weight/gyro to the point it is fixed, in this case your hand) does vary. ̉ۡ, which I assume you interpret as the angular velocity of the mass around your hand, can vary too.

[Momentus]

The equation was specifically only valid under these conditions:

You have a gyro precessing with angular velocity ̉ۡ. There is no such thing in your setup.

z is small compared to L. Not the case in your experiment, they are about equal.

The formula is for the movement component induced by the precession. In your experiment, you are using the other, regular pendulum motion mode.

So yeah, of course the formula would describe the experiment incorrectly. The experiment does not even come close to testing the formula.

Furthermore, contrary to your claims, L (the RELEVANT rope length.Total rope length does not matter, what does matter is the length from the weight/gyro to the point it is fixed, in this case your hand) does vary. ̉ۡ, which I assume you interpret as the angular velocity of the mass around your hand, can vary too.

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.

"only valid under these conditions: You have a gyro precessing" does that mean it is not an orbiting mass? In the Laithwaite ice tower, did a gyro behave differently from an inert mass?

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.

Momentus