[Momentus (OP)]

This is a machine. It is indisputable that by judicious manipulation of the two motors the gyroscopes can be spun up to speed about the yellow axis and set freewheeling and the frame can be rotated about the red axis.

Now take one of the gyroscopes, it is rotating about two separate axes. The yellow axis, and the red axis at the same time. The angular velocity diagram shows these two simultaneous rotations.
A similar diagram is drawn for the other contra rotating gyroscope.

Assuming a spherical gyroscope rotor the polar moment of inertia will be the same for spin about OxA, OxC, OxD and OxB as they all pass through the centre of the sphere.

If Ω is the spin speed of the gyroscope at the start, then the angular momentum is I Ω

Precessing a gyroscope does not change the magnitude of its angular momentum. The torque is reacted by the changing direction of the angular momentum.

The direction of the angular momentum is the vector sum of the two rotations. The magnitude has not been altered by the application of the torque at the blue axis, only the direction.

Thus in the diagram OxC shows the angular momentum as the frame precesses about the red axis at low speed.

OxD is the point at which speed of precession is equal to spin sped and is the limit where increasing the precession speed will change the direction of the resultant angular momentum.

OxE shows the angular momentum as the frame precesses about the yellow axis at low speed.




Diagram of Angular momentum
By manipulating the speed of the two motors, the direction of the angular momentum can be moved through 90 degrees.

The vector sum of the initial angular momentum of the system is null.

The vector sum of the final angular momentum of the system is positive.

Further posting will follow. I am off now to watch a recording of the Northampton Saints Rugby match. Come on you Saints.

[Halc]

If a gyroscope is rotating about any axis, other than the spin axis, then that is precession and Orthogonal to that axis is the couple.

I'm fine with those terms. The term couple is not always used, but yes, it seems always orthogonal to the precession axis.  It is also orthogonal to the spin axis in your model, but that is just because you set it up that way.  It need not be at all.

For clarity, let the red axis be along the red flywheel axis, The yellow axis is the gyro axis and the blue axis is orthogonal to these. In the model the red axis does not move, the yellow and the blue axes move but remain orthogonal to red.

These differ from the picture in the OP, but I like these better.
The description is incomplete.  Which way is positive for each?  A coordinate system is less useful if no designation is made as to which way is positive.
I'll choose if you don't mind:  Red is positive to the right where the flywheel is.  Yellow is positive to the upper-right (as per OP picture).  The yellow gyro near to the flywheel (which I will call gN) has positive spin, and the far gyro (gF) has negative spin.  As for the couple, positive is up when the spin is oriented as per the picture.  Yes, both these two precess around, so not permanently in any one direction.

In the model we have forced precession along the red axis, with equal and opposite torque about the blue axis and spin about the yellow axis. This is the starting condition.

You need to identify the components so we can talk about them clearly.  Each yellow wheel is a gyro, designated gF and gN respectively.  The device is in two halves which consists of the assembly A and the red flywheel R, both of which spin along the red axis, with say R having positive spin and A having negative spin.  Finally there is the entire system S which is A and R combined.  S is suspended in a 3-gimbal that allows free rotation in any axis, and thus prevents any external torque from being applied to S though the bearings that hold the main shaft. The angular momentum of S thus always remains zero, and the device stays put.

Part of the problem with this post is the absence of identification of component being discussed.

The spinning gyroscope rotor is forced to precess about the red axis as that is the only axis that permits rotation

Each gyro gF and gN is now free to precesss any way physics says it will.  I've purposely freed the system from any restraints that prevent motion in directions other than the red axis.  So if it doesn't rotate, then it is because nothing is attempting to rotate on the other axes, not because of any restraints.  The scenario hasn't changed since I assert that no such motion will occur.  If a gryo precesses, then there must be a torque being applied to it along the couple, which is not the red axis since that's the precession axis.

This simultaneously induces a torque about the other orthogonal axis, the Blue axis.

Yes, that's the only torque, the reaction torque to the torque being applied to each gyro. Blue is the couple.

When the frame containing the spinning gyroscopes is rotated by the red motor,

This is what I'm calling assembly 'A'.  The red motor only serves to change the angular momentum of A from zero to nonzero. We can assume that the moment of A is the same as the moment of R (whether the gyros are spinning or not), and thus they spin at equal but opposite directions, all without help of the motor which is only used to change the spin, not to maintain it. So in steady state, there is no torque exerted along the red axis.

That rotation induces an orthogonal couple at the blue axis.

It induces an orthogonal couple on gN and gF, but not on A.  This is what I mean by you switching objects without identifying them. Label your objects, or your statement becomes meaningless. The spin on A along the red axis does not induce an orthogonal couple on A because there is no torque on the red axis and thus no precession of A. Your statement seems to imply there is one, but lacking precise language, it is simply ambiguous.

As the spin of the gyroscopes are equal and opposite, these torques, induced by the precession about the red axis are equal and opposite.

Fine.  This statement identifies the gyros gN and gF, and not the frame A. I agree with this.

My conclusion.

The conditions for the flywheels to act as gyroscopes are fulfilled. A spin axis of the rotor, Yellow, A torque axis, Blue and a precession axis, Red.

We're all still wondering what 'anomalous motion' is being asserted on top of this description, with which I do not disagree.  The thread is entitled 'dark motion', yet no dark motion has been identified.

This is a machine. It is indisputable that by judicious manipulation of the two motors the gyroscopes can be spun up to speed about the yellow axis and set freewheeling and the frame can be rotated about the red axis.

With the frame also set freewheeling, yes. In a frictionless environment, all motors serve only to change the spin rates, not to maintain them.

Now take one of the gyroscopes, it is rotating about two separate axes. The yellow axis, and the red axis at the same time.

No, it is rotating only about yellow axis.  Precession is not rotation. The former is the product of the angular momentum of the gyro.  The latter is the product of a change in angular momentum, and that requires continuous external application of torque.

The angular velocity diagram shows these two simultaneous rotations.

A spin vector and a precession vector are thus different units and cannot be directly added or compared as you are doing in your diagram. The black vectors are meaningless. Your diagram also shows a dotted circle of constant magnitude which seems to imply that the magnitude of the precession angular velocity must be identical to the spin angular velocity, which is untrue.  The precession angular velocity is purely a function (in this case) of how much spin the red motor happens to put on A, which you describe below as 'low speed', contradicting your diagram.

If Ω is the spin speed of the gyroscope at the start, then the angular momentum is I Ω

Ah good.  A number.  If gN has an angular momentum of I Ω, then gF has an angular momentum of -I Ω. Let's say the frame assembly A has an angular moment of 3 (two for the pair of spheres, and 1 more for the frame mechanism), and the red flywheel thus also has a moment of 3. We've not designated a symbol for the angular velocity of R or A.

The direction of the angular momentum is the vector sum of the two rotations.

The direction the angular momentum is by definition that of the one spin rotation, so this statement is wrong. You cannot add a momentum vector to a torque/precession vector any more than it is meaningful to add velocity to acceleration. Different units.

The rest of your post seems to depend on this invalid mathematics, and is thus wrong.

The vector sum of the initial angular momentum of the system is null.

There is only one system, and its angular momentum is always zero, and there are no other vectors to add to it. It would be more correct to say that the vector sum of the angular momentum of the various components of the system S is null. The components definitely do have angular momentum, but since S is an isolated system, the angular moment of S cannot change from zero.

The vector sum of the final angular momentum of the system is positive.

Well, given all the invalid addition of vectors of different units, perhaps you came to that conclusion.  If so, positive in which direction, and you need to show this without the invalid mathematics.  Only add vectors of the same units.

[Miklos]

Dear Halc,

Could you please further elaborate on your statement below:

Precession is not rotation.

I would say, precession is a process, that manifests itself as a rotation. Vaguely formulated.
At the end one will see a rotation about an axis different than the axis of spin.

[Momentus (OP)]

With a light frame and heavy spherical gyroscopes spinning in equal and opposite directions

Scenario 1)

Viewed as a black box, it will behave like a rock with the same mass.  Net angular momentum is zero.

Start the motor on the red axis.

. So the red wheel spins one way, and the contraption to the left spins the other way.  Still zero net angular momentum.

Nope. At the end, one will see a continuous change in the axis of spin, which is mandated by the change in the angular momentum resulting from the continuous applied torque.  But the gyro always has only one axis of spin, and one rate at which it is spinning on that axis. If at any point in time the torque is removed, it will remain spinning only about that one axis.

So let us do just that. Remove the torque. Stop the rotation of the yellow mass about the red axis.

The Red flywheel continues to rotate. The contraption does not.

[Momentus (OP)]

Scenario 2

1. Run the gyroscope motor to bring the gyroscope rotors up to a modest speed. As the rotors are are touching, they counter rotate. Equal and opposite. Now let the motor freewheel.

2. Run the reaction motor. This rotates the Gyroscopes about their precession axis, accompanied by equal and opposite torque orthogonal to the spin axis. There can be no torque exerted by the reaction motor along a precession axis, therefore no substantial movement of the reaction flywheel.

Remove the Torque. Gyroscopes stop precessing. Back to square one.

[Halc]

So let us do just that. Remove the torque. Stop the rotation of the yellow mass about the red axis.

The Red flywheel continues to rotate. The contraption does not.

This comment shows a complete lack of understanding of rotation and momentum.
If assembly A (what you're calling the yellow mass) is spinning about the red axis, then its angular momentum vector is along said red axis, and with any external torque removed, it must continue to spin, not stop (in complete violation of conservation laws). Your assertion otherwise shows said lack of understanding. Its angular momentum cannot just vanish. It's a conserved quantity. It has to go somewhere.
It would require external torque on it to stop it, and of course this would also stop the red wheel, which always has equal and opposite angular momentum of assembly A.

Scenario 2

1. Run the gyroscope motor to bring the gyroscope rotors up to a modest speed. As the rotors are are touching, they counter rotate. Equal and opposite. Now let the motor freewheel.

2. Run the reaction motor. This rotates the Gyroscopes about their precession axis, accompanied by equal and opposite torque orthogonal to the spin axis. There can be no torque exerted by the reaction motor along a precession axis, therefore no substantial movement of the reaction flywheel.

Remove the Torque. Gyroscopes stop precessing.

Unclear description.  Remove what torque? The torque from the reaction motor isn't attached to any gyroscope, it is attached to assembly A and the wheel R, so it gives those two things equal and opposite angular momentum, and thus both must spin, contradicting your assertion above which is in direct violation of Newton's 3rd law.
That motor applies no torque to either gyroscope gN and gF, as evidenced by the fact that neither precesses around an axis perpendicular to the torque along the red axis.  The reaction motor thus exerts torque only against the identical (arbitrary designation) moments of the wheel R and assembly A, which is the same whether the gyros within are spinning or not since the angular momentum of A is unaffected by the gyros spinning or not.

Yes, there is torque exerted on each gyro, but not along the red axis, so the reaction motor does not contribute to it.

[Momentus (OP)]

 Forced Precession. Or Induced torque.

Tb=JΩyΩr

Tb is the torque at the blue axis in Nm 
Ωy is the angular velocity about the yellow axis in rad/sec Gyroscope spin
Ωr is the angular velocity about the red axis in rad/sec Precession


Consider one gyroscope spinning about the yellow axis. Apply a torque at the blue axis, it will rotate about the red axis in a clockwise direction.

Consider the other gyroscope spinning about the yellow axis in the opposite direction. Apply a torque at the blue axis, again in the opposite direction, it will rotate about the red axis in a clockwise direction.
 
Put the gyroscopes in a frame and repeat the exercise, applying torque at the blue axis etc. They will rotate about the red axis in perfect harmony, precessed by the blue opposite torques. The Red flywheel plays no part, remains unmoved. The net angular momentum has not changed, it is still null

Everybody happy? Take a video. Go to lunch

Next spin gyroscopes as before and rotate them about the red axis, using the Red Reaction flywheel, at exactly the same precession speed as before. The net angular momentum has not changed, it is still null. Take a video. You already had lunch.

What is the physical difference in the speed and torque between the two cases? What motion or force is there in the first case that is not identical in the second case?

There is no difference. The speeds of the gyroscopes are the same, the torques are the same and the precession speed is the same.

Shown the video, how to tell the difference? So what did the red flywheel do? There is no difference in the net angular momentum of the system in either case, so it did the same thing in both cases, nothing.

Run the reaction motor. This rotates the Gyroscopes about their precession axis, accompanied by equal and opposite torque orthogonal to the spin axis. There can be no torque exerted by the reaction motor along a precession axis, therefore no substantial movement of the reaction flywheel.

[Momentus (OP)]

is comment shows a complete lack of understanding of rotation and momentum.
If assembly A (what you're calling the yellow mass) is spinning about the red axis, then its angular momentum vector is along said red axis, and with any external torque removed, it must continue to spin, not stop (in complete violation of conservation laws). Your assertion otherwise shows said lack of understanding. Its angular momentum cannot just vanish. It's a conserved quantity. It has to go somewhere.

I am pointing out the inconsistency I see in your definition of precession as being a different form of rotation.

Quoting Conservancy laws. You have missed the entire point of my thread.

If I may try to see what you are saying. I observe the contraption rotating about the red axis, it is either spinning or precessing and these are different.

 If it is precessing that is because I am exerting equal and opposite torques about the blue axis. It is rotating about the red axis and that is precession.

If it is spinning, that is because the equal and opposite torques are being generated by rotating it about the red axis with an external torque. It is rotating about the red axis and that is spinning.

No I have got that wrong. There cannot be an external torque spinning it, that would increase the spin speed over time. Inertia?

To calculate Gyroscope torque There is a formula.  The angular velocity of the Gyroscope is given as rads/sec. The precession is also given in rads/sec. Same units? Can multiply spin by precession? Maybe even vector sum angular velocity when expressed as rads/sec?

[Miklos]

Dear Halc,

for me, rotation is a kinematic quantity, it is defined without forces and torques. Let’s say, that rotation is a transformation, or a series of transformations.  It doesn’t matter, whether it is a steady or constantly changing rotation, both could be described as a series of successive rotations about some steady or instantaneous axis. 

The precessional motion of a gyro can be quite steady. In this case it looks like steady rotation, and it is a rotation, it has angular velocity and a moment of inertia, therefore it has angular momentum too. Although it can be quite small in some cases. 

You use in your definition for rotation the notion “external torque”. I assume, that you do not consider gyroscopic moments as external. And to some degree I would agree. 

But if one considers the equation of motion of a precessing (and maybe nutating) gyroscope, one will find some terms called inertia quadratic velocities. If one starts with the Lagrangian, then these terms are derived from the kinetic energy. They are torque-like quantities (gyroscopic moments), in case of rotations as generalized coordinates. To be able to solve the equations of motion, they should be moved to the right hand side along with the generalized external forces. They are not external, but they act on the system, as they were. 

The distinction of let's say "standard" rotation and "precessional" rotation is somewhat exotic to me. Could you give me some literature references, where this is discussed in detail?

Thanks, and sorry about being off topic so long. 

Best regards,
Miklos

[Halc]

Forced Precession. Or Induced torque.

Tb=JΩyΩr

Tb is the torque at the blue axis in Nm 
Ωy is the angular velocity about the yellow axis in rad/sec Gyroscope spin
Ωr is the angular velocity about the red axis in rad/sec Precession

Exactly so. This is what I was talking about here:

Yes, there is torque exerted on each gyro, but not along the red axis

Yes, Tb on one gyro (gN say) is significant, probably much larger magnitude than what either motor is putting out, especially in the steady state where neither motor is exerting any torque at all. Where is the reaction for that torque? Our assembly A doesn’t start rotating about any blue axis even though it is free to.

Consider one gyroscope spinning about the yellow axis. Apply a torque at the blue axis, it will rotate about the red axis in a clockwise direction.

Clockwise isn’t a direction without specification of a PoV.  Your statement lacks directions for all three vectors. Be precise.
Anyway, yes, there is a torque about the blue axis and thus it precesses (or rotates, if you ask Miklos) about the red axis, exactly as we observe.

Consider the other gyroscope spinning about the yellow axis in the opposite direction. Apply a torque at the blue axis, again in the opposite direction, it will rotate about the red axis in a clockwise direction.

Very good. Equal and opposite torque about the couple, so no net torque on assembly A.  This is why it has no more resistance to rotation with the gyros spinning or not.  You’d not be able to tell from holding it if they gyros were spinning. Listening would be your best bet.

Put the gyroscopes in a frame and repeat the exercise, applying torque at the blue axis etc. They will rotate about the red axis in perfect harmony, precessed by the blue opposite torques. The Red flywheel plays no part, remains unmoved.

The red wheel doesn’t remain unmoved because something had to get assembly A spinning in the first place. Without that, there’s no precession of the gyros and no blue torque. But such motion is real rotation, which continues once torque is removed. Precession doesn’t do that, so assembly A spins and has angular momentum about the red axis, and the red wheel has equal and opposite angular momentum about the same axis.

The net angular momentum has not changed, it is still null

You’re describing a system with nonzero angular momentum. This cannot be since no torque has exerted from outside the system.

Next spin gyroscopes as before and rotate them about the red axis, using the Red Reaction flywheel, at exactly the same precession speed as before. The net angular momentum has not changed, it is still null.

This time I agree that the net angular momentum is null, as it always has been.

What is the physical difference in the speed and torque between the two cases? What motion or force is there in the first case that is not identical in the second case?

The first case has external angular momentum used to spin assembly A, but not spin wheel R.  Since external momentum was introduced, it has nonzero system angular momentum. That’s the difference.

There is no difference. The speeds of the gyroscopes are the same, the torques are the same and the precession speed is the same.

The system angular momentum is not the same.

Shown the video, how to tell the difference?

One has the red wheel spinning, the other not. Not sure why a video is suddenly needed. You seem to have no such actual device built, although it seems simple enough to make one.

So what did the red flywheel do? There is no difference in the net angular momentum of the system in either case, so it did the same thing in both cases, nothing.

If you think two systems, that are identical except for the red wheel spinning in one and not the other, have the same angular momentum, then you’re just lying to yourself. Surely the red wheel contributes to angular momentum of the system of which it is a part.

I am pointing out the inconsistency I see in your definition of precession as being a different form of rotation.

If I may try to see what you are saying. I observe the contraption rotating about the red axis, it is either spinning or precessing and these are different.

The ‘contraption’ I think is what I call assembly A. Yes, I consider assembly A to be rotating (spinning) about the red axis, all without torque required to maintain that spin.  The gyros on the other hand are precessing since their spins are about the yellow axis, and require continuous torque to maintain that precession. You seem to describe all that pretty well in the above post, yet somewhere you get lost with what goes on along the red axis.

If it is precessing that is because I am exerting equal and opposite torques about the blue axis.

There is no torque on A, so no, I don’t consider A to be precessing. Torque is only needed to get it spinning from a stop, and opposite torque can be used to stop it again.  The red wheel always has equal and opposite angular momentum to A.

If it is spinning, that is because the equal and opposite torques are being generated by rotating it about the red axis with an external torque.

No external torque is needed for A to continue spinning any more than the red wheel needs continuous torque to continue spinning.

No I have got that wrong. There cannot be an external torque spinning it, that would increase the spin speed over time. Inertia?

Assembly A has a moment (which I specified as 3, same as the red wheel), so yes, continued torque along the red axis would make it spin ever faster. Earth’s spin is continuously slowing because there is always external torque on it.  The day used to be like a 3rd of what is now.

To calculate Gyroscope torque There is a formula.  The angular velocity of the Gyroscope is given as rads/sec. The precession is also given in rads/sec. Same units?

Both are rads per second if you look at it like that, but adding them does not give a single vector that describes the motion of the thing. One is an angular velocity vector, and the other the angular velocity of the angular velocity vector, essentially the rate of direction change of the first vector. One is the derivative of the other. No, they’re not the same units.

You also had that funny curved dotted line, which is not how vector addition works if the vectors actually had been the same units.

Can multiply spin by precession?

Multiply, yes. It was adding apples to oranges that doesn’t work. Your formula at the top of this post multiplies them, yielding torque.

for me, rotation is a kinematic quantity, it is defined without forces and torques. Let’s say, that rotation is a transformation, or a series of transformations.  It doesn’t matter, whether it is a steady or constantly changing rotation, both could be described as a series of successive rotations about some steady or instantaneous axis.

Fine, you have different definitions. I admit I’m not speaking on any authority for the terms. I’m just trying to distinguish the one kind of motion from the other.

Your definition violates Newtons first law of rotational motion: "1) a rotating body will continue to turn about its axis of rotation with constant angular momentum, unless an external couple or eccentric force is exerted upon it" https://www.asu.edu/courses/kin335/documents/Angular%20Kinetics%20and%20Angular%20Momentum.pdf
Precession about an axis will cease in the absence of external forces, so you calling this rotation seems contrary to the prevailing definition.
I found the law on http://www.4physics.com/phy_demo/newton/newton_rot2.htm as well, but it is clearly wrong: "The rotational principle of inertia: In the absence of a net applied torque, the angular velocity remains unchanged."  That says that the skater will not spin faster when she pulls in her arms since no net torque is being applied. So just because it's out there doesn't mean it's correct.

The precessional motion of a gyro can be quite steady. In this case it looks like steady rotation, and it is a rotation, it has angular velocity and a moment of inertia, therefore it has angular momentum too.

That angular momentum isn’t going to maintain the precession in the absence of continuous torque, so I balk at using such terms.  Our assembly A on the other hand has angular momentum, and it rotates slow or fast at our choosing, and will continue to rotate without external torque, so that’s spin. Assembly A does not precess.

The distinction of let's say "standard" rotation and "precessional" rotation is somewhat exotic to me. Could you give me some literature references, where this is discussed in detail?

Sorry. I’m not working off any literature, just the mathematics of simple situations and conservation laws.

[Momentus (OP)]

Quote from: Momentus on Yesterday at 15:25:18
Forced Precession. Or Induced torque.

Tb=JΩyΩr

Tb is the torque at the blue axis in Nm 
Ωy is the angular velocity about the yellow axis in rad/sec Gyroscope spin
Ωr is the angular velocity about the red axis in rad/sec Precession

Exactly so. This is what I was talking about here:
Quote from: Halc
Yes, there is torque exerted on each gyro, but not along the red axis
Yes, Tb on one gyro (gN say) is significant, probably much larger magnitude than what either motor is putting out, especially in the steady state where neither motor is exerting any torque at all. Where is the reaction for that torque? Our assembly A doesn’t start rotating about any blue axis even though it is free to.

You ask where is the reaction to the torque applied at the blue axis of the gyroscope which is spinning about the yellow axis. You ask because you do not know?
Then you say it does not start rotating about any blue axis. It is a gyroscope. Why do you say that a gyroscope does not rotate about its torque axis? Did you expect it to?
You do not say how the torque is reacted. Again is that because you do not know?
We should be told. The nation demands an answer, How is the torque reacted?

 

Clockwise isn’t a direction without specification of a PoV.  Your statement lacks directions for all three vectors. Be precise.
Anyway, yes, there is a torque about the blue axis and thus it precesses (or rotates, if you ask Miklos) about the red axis, exactly as we observe.

I stand corrected. But pleased that you managed to solve the difficult question of POV without my help.

You also understand that there is clockwise movement around the red axis. You accept that the second gyroscope also moves in the same way, about the same axis. Do follow my logic, that if you have two gyroscopes moving in the same direction you would see them moving in the same direction?

Very good. Equal and opposite torque about the couple, so no net torque on assembly A.  This is why it has no more resistance to rotation with the gyros spinning or not.  You’d not be able to tell from holding it if they gyros were spinning. Listening would be your best bet.

And that because they are moving in the same direction, you would not need to listen, you would see the difference?

The red wheel doesn’t remain unmoved because something had to get assembly A spinning in the first place. Without that, there’s no precession of the gyros and no blue torque. But such motion is real rotation, which continues once torque is removed. Precession doesn’t do that, so assembly A spins and has angular momentum about the red axis, and the red wheel has equal and opposite angular momentum about the same axis.

You defined assembly A.

The device is in two halves which consists of the assembly A and the red flywheel R, both of which spin along the red axis, with say R having positive spin and A having negative spin. 

With a light frame and heavy spherical gyroscopes spinning in equal and opposite directions

That is component A.

The two gyroscope are both moving in the same  direction, reacting the equal and opposite torques applied at the blue axis. You have agreed that this is the case. The light frame is attached to the gyroscopes and thus rotates with them all driven by the torque applied at the blue axis. It is an applied torque. It is reacted by orthogonal rotation.

So it has been established and you have agreed that something has got A moving. That observed motion is the reaction to the applied torque. Applied torque, equal and opposite applied torque. Applied at the blue axis, precessing about the red axis.

The red wheel is not needed, it has nothing to add.

[Colin2B]

The distinction of let's say "standard" rotation and "precessional" rotation is somewhat exotic to me. Could you give me some literature references, where this is discussed in detail?

I think halc is a little busy on a number of threads and we are getting a lot of spam which takes up our time.
I’m not aware of any detailed discussion in the literature, most I’ve seen on definitions tends to be in passing. For gyroscopes spin is defined, but precession is often defined as a change in inclination of the spin axis which probably covers most situations. Generally, precession of a gyroscope is taken to mean the response to an input rotation or a torque/couple/moment, but the OP is using it here to refer to the input (rather than the gyroscopic precession) without qualifying the term eg input precession. This is leading to some confusing statements and lack of clear thinking.

To explain more, what halc has designated assembly A is surprisingly representative of a model of a boat antiroll system I helped build many years ago as a lab demo. The workshop team built a model ship’s hull so it could be used in a wave tank, but I’ll describe it relative to assembly A to make it clear what I mean about the precession confusion.
If you consider the assembly A with its 2 yellow flywheels as the side view of a boat (I’ll leave you to imagine pointy end) with the yellow flywheel axles vertical, we have an illustration of a real life gyro stabiliser used in luxury yachts to reduce rolling at anchor (rolling is a nauseating motion far worse than pitching). Usually there is just one flywheel, but 2 can be used, and in this application the flywheel is on a gimble and allowed to pitch forwards and backwards around the blue axis.
In this system the red axis plays the part of the rolling motion, waves broadside will rotate the (yellow) flywheel axle around the red axis (ok, input precession) and the flywheel will respond by trying to pitch - gyroscopic precession- around the blue axis in either direction depending on the direction of rotation, thus the precession axis is across the boat - side to side (perpendicular to the plane of the frame in assembly A, blue axis). Of course in the OP’s model the blue axis is locked to the frame and cannot pitch, although it tries to with the frame taking the strain. In the real world system the precession axis is provided with a brake, if the brake is fully released the flywheel will pitch (precess) freely and there is a resulting torque that directly opposes the rolling motion (red axis), however it is almost instantaneous and of such force that is can damage the bearings and gyro mountings. If the brake is locked the flywheel cannot precess and there is no roll-opposing torque. So in the real world system the brake is set to provide controlled precession by applying a negative torque to the precession axis, usually under electronic control to allow adjustment. In the model I built there was no electronic control, just a friction brake to demonstrate the ability to control the roll-opposing torque and the effect of locking the precession axis to prevent any counter torque. In theory you could also apply a positive torque to the gyroscopic precession axis in order to increase speed up the precession and increase the roll-opposing torque.
 Although you can use 2 flywheels, in the OP’s  model the axles of both of these are locked to the frame and unable to precess, so no roll-opposing torque on the red axis results. Because the gyroscopic precession axis is locked it doesn't matter whether the flywheels are spinning or not, whether there is 1 or 2, or whether they are co-rotating or counterrotating.
The OP claims anomalous motion, but it is not clear what this is. He has not provided videos or even photos, so there is some doubt that the model exists other than as a drawing.
If it does, item 4 of the opening post is telling:
“ 4. As the spin axis of the gyroscopes is now aligned with the Reaction motor, precession has changed to what was the spin axis and the gyroscope motor can be used to brake the precession to a halt.”
I wonder if he is thinking that stopping the yellow gyroscopes is stopping what he has now defined as the precession axis, even though nothing has changed in reality. In other words, this is all in the mind and a question of definitions rather than a physical change.
To be honest, I’m surprised the yellow flywheels do much freewheeling as the rolling friction, drivebelt friction and motor friction (including magnet effect) must be quite large.
I did ask (no reply) how the reaction motor is connected to the reaction flywheel as this probably acts as an automatic transmission due to friction and magnetic effects. This is why in most lab demos the spinup motors are external to the system and can be disconnected rather than influence the experiment.

I gather that you are generally skeptical, but hopeful 

[Momentus (OP)]

For gyroscopes spin is defined, but precession is often defined as a change in inclination of the spin axis which probably covers most situations.

This change in inclination can be seen in my velocity diagram as the vector OC of the gyroscope when it is precessing, displaced from the vector OY when the gyroscope is not precessing.

Generally, precession of a gyroscope is taken to mean the response to an input rotation or a torque/couple/moment,

,

There is no input rotation at the torque axis, only a couple, torque, or moment. It is not possible to rotate a gyroscope about the torque axis. Rotation of a gyroscope, (other than spin) is precession of the gyroscope. The rotation is orthogonal. There is no other mode of rotation.
My assumption was that these facts were general knowledge, as such any reference made to rotation (other than about the spin axis) is a reference to precession and vice versa.

 Bear in mind that the OP is a novice and was labouring under the delusion that a forum god would know the basics.

but the OP is using it here to refer to the input (rather than the gyroscopic precession) without qualifying the term eg input precession. This is leading to some confusing statements and lack of clear thinking.

Gyroscope basics.
A gyroscope reacts a torque with orthogonal motion.
A gyroscope reacts motion with an orthogonal torque.

My use of the word precession is motion orthogonal to torque and spin.

Your work on gyroscope stabilization of boats has Victorian precedents https://en.wikipedia.org/wiki/Anti-rolling
 

In this system the red axis plays the part of the rolling motion, waves broadside will rotate the (yellow) flywheel axle around the red axis (ok, input precession) and the flywheel will respond by trying to pitch - gyroscopic precession- around the blue axis in either direction depending on the direction of rotation, thus the precession axis is across the boat - side to side (perpendicular to the plane of the frame in assembly A, blue axis).

The waves do not rotate the boat about the red axis. They may well try to do so, but the gyroscope will not allow the motion. The waves apply a torque.

In the real world system the precession axis is provided with a brake, if the brake is fully released the flywheel will pitch (precess) freely and there is a resulting torque that directly opposes the rolling motion (red axis), however it is almost instantaneous and of such force that is can damage the bearings and gyro mountings. If the brake is locked the flywheel cannot precess and there is no roll-opposing torque. So in the real world system the brake is set to provide controlled precession by applying a negative torque to the precession axis,

The your “negative torque at the blue axis” means that the system is subjected to two torques. The original roll torque from wave action and the brake torque. These are vector quantities, and can be summed to produce a resultant torque vector. This torque precesses the gyroscope orthogonally. This resultant precession vector can in turn be expressed as vectors around the red and blue axis. There is rotation about the red and blue axes.

Although you can use 2 flywheels, in the OP’s  model the axles of both of these are locked to the frame and unable to precess,

That should read “unable to precess about the blue axis”. For clarity.

It is free to precess about the red axis. When it is precessing about the red axis there is a torque reacting that motion about the blue axis. May I repeat that for emphasis? Reacted by a torque at the blue axis. The gyroscope reacts the torque orthogonally.

If the flywheels ARE Spinning about the yellow axis
The rotation about the red axis is gyroscopic precession, and it is reacted by an orthogonal torque at the blue axis.
If the flywheels ARE NOT spinning about the yellow axis, the reaction to the rotation of the red axis is torque about the red axis.  Or as Newton might have said had he been asked “along the line of action”

With the wheels spinning force is changing the momentum vector’s direction You called that “inclining the spin angle“. Precession, orthogonal.

When there are different conditions, when the wheels are not spinning. Then the magnitude of the momentum is changed, along the right line, in the direction of the angular velocity.

 Although you can use 2 flywheels, in the OP’s  model the axles of both of these are locked to the frame and unable to precess, so no roll-opposing torque on the red axis results. Because the gyroscopic precession axis is locked it doesn't matter whether the flywheels are spinning or not, whether there is 1 or 2, or whether they are co-rotating or counterrotating.

It does matter.

Static Flywheels
The inertia, the momentum change, of the static flywheels must be overcome by a torque along the line of action of the resultant rotation. The angular velocity vector is extended in length.
Spinning flywheels
The inertia, the momentum change, of the spinning flywheels must be overcome by a torque orthogonal to the line of action of the resultant rotation. The angular velocity vector is changed in direction.

There is a difference, the difference matters.

The OP claims anomalous motion, but it is not clear what this is. He has not provided videos or even photos, so there is some doubt that the model exists other than as a drawing.

The model as drawn does not exist. I felt that it gave enough information for any interested party to construct one. I asked that some one might do so.

The model I did make and patented back in the day was designed as an automotive transmission. It is an order of magnitude more complicated. Photos and Videos would further confuse a complicated subject.

You wrote your post in response to @Miklos

Quote from: Miklos on 15/09/2020 18:16:23
The distinction of let's say "standard" rotation and "precessional" rotation is somewhat exotic to me. Could you give me some literature references, where this is discussed in detail?

Whilst your post was both interesting and instructive, I don't see how it addressed "standard" rotation and "precessional" rotation, The question asked by @Miklos
.

[Momentus (OP)]

Now to consider the vectors
A particle of mass dm has velocity V relative to point O
A continuous force Fc is exerted at a right angles to the direction of travel (centripetal force) and the path of dm is changed to orbit around the point O with an angular velocity of Ωdm . If this force is removed then dm will move in a straight line, at the same speed as it had before. Force at right angles to the direction of travel changes momentum by changing direction, not by changing magnitude.

An orthogonal torque is applied, which exerts a variable force Ft with a max value as dm passes the north and south poles of its orbit and zero at the equator.

The resultant vector sum of Fc & Ft is Fx and is also orthogonal and determines the path followed by dm. If Fx is removed then dm will move in a straight line, at the same speed as it had before.

Rotation has angular momentum to it, and obeys the angular version of Newton's first law: A rotating object will continue rotating in the absence of an external force. This isn't true of precession, which ceases upon removal of external force (torque). Hence my saying that precession is not rotation. It is far more akin to acceleration since it is a process resulting from a force (torque).

The particle dm is part of a rigid mass. The mathematical determination of the path of dm whilst acted upon by Fx is complex and can be found in a gyro – dynamics text if you wish to know more.
Suffice to say that it resolves to the formula

                                             Tb=JΩdmΩr

The actual path of the particle dm can be traced by observing a wheel that is rotating about the axes  y and r. I did this by using the stop motion technique. I used a polystyrene ball and a felt marker pen to make a dot on the surface of the sphere, moved it through a small arc in one direction Sr, then a larger arc in the other Sy and made a second mark and so on tracing a curving path on the surface of the ball.

Sy is the displacement about the yellow axis. Sr is the displacement about the red axis

Their vector sum, the arc Sdm can be resolved to the tangential speed of the Gyroscope, and hence to the speed of rotation, Ωdm This value does not change as it is derived from the constant instantaneous speed of dm, acted upon by Fx.

Sy can be resolved to an angular velocity Ωy. Sr can be resolved to an angular velocity Ωr 

The interaction of the angular velocity of Ωy  and the angular velocity of Ωr are shown in Angular momentum diagram 2

As the gyroscope rotor under consideration is a sphere, the polar moment of inertia is the same for any plane intersecting the sphere. As such Diagram 2. can be interpreted as a momentum diagram.

The equal and opposite momentum generated by the contra rotation of the gyroscopes is turned through 90 degrees in reaction to equal and opposite torque, resulting in the positive momentum at the red axis, the vector sum of the original momentum about the yellow axis.

[Momentus (OP)]

There is always the other way to operate the device.
In the OP the process started by contra-rotating the two gyroscopes about their yellow axis.
This time, do something different, start the motor attached to the red reaction flywheel which rotates the flywheel about the red axis and counter rotates the two gyroscopes, also about the red axis. The gyroscopes are both spinning in the same direction. The sum of their angular momentum is equal and opposite to that of the flywheel. This generates momentum in equal and opposite quantities. You could if you wanted to, detach the flywheel, take it away and use the momentum that has been generated to tighten a bolt.
Equal and opposite momentum does not mean no momentum. But to be clear do not do that yet. Leave it where it is.

Now for the difficult to understand part.
Uncouple the red flywheel. The two heavy gyroscopes and the light frame will continue to spin about the red axis. The flywheel will continue to spin.

Start the yellow gyroscope motor to counter rotate the rotors about the yellow axis.

This movement, the physical rotation by the drive belt of the spherical rotors about the yellow axis is orthogonal to the spin of the gyroscopes which is about the red axis.

Yes very difficult to visualise. But possible if you try.

 The two gyroscope rotors are spinning about the red axis and therefore the orthogonal rotation about the yellow axis is, by definition precession.

The dynamics of a gyroscope dictate that there is a torque present at the blue axis, again by definition. Each of the rotors will generate a torque. The torques will be equal and opposite..

When a rotor spinning about an axis(red) is subjected to a torque about an orthogonal axis (blue) there is precession about the other orthogonal axis (yellow)

Remember, the red wheel is disconnected, it has no influence.

The condition as described above is best represented by the angular velocity vectors in angular momentum diagram 2
OB is the Spin axis
OA is the precession axis.

OE is the vector sum of the actual physical rotation of the rotor about the yellow axis and the actual in the real world physical rotation of the rotors about the red axis.

The original spin vector OB is displaced by the orthogonal force exerted by the torque at the blue axis This changes only the direction of the vector, it does not change the magnitude. The length of the line representing the vector does not change. Hence the dotted curved line which Halc queried.

 

[Halc]

Bear in mind that the OP is a novice and was labouring under the delusion that a forum god would know the basics.

The OP may then stand corrected in this delusion.  The rank is assigned by the site software and is purely due to the number of posts (10000 in this case) and nothing else. There are some here who easily have posted more than that and are 'forum god's and know little of said basics. Colin is not one of those.
There are better clues as to which posters are likely to know what they're talking about. Think about it.

You seem to be posting for our educational benefit, and have asserted no more of this 'anomalous dark motion', so I've pretty much been leaving it alone.  But I've accumulated a few comments:

A particle of mass dm has velocity V relative to point O

Relative to frame F, which defines a spatial point O.

A continuous force Fc is exerted at a right angles to the direction of travel (centripetal force) and the path of dm is changed to orbit around the point O with an angular velocity of Ωdm

This makes a lot of assumptions, primarily that Fc is not continuous but actually changing over time.  A continuous force on a particle actually causes a more or less parabolic trajectory.
The statement makes the assumption that the particle is in the vicinity of point O.  That point could be on the other side of the galaxy, and a force Fc exerted at right angles to its motion (in frame F) is hardly going to put it in orbit about O.

Just pointing out that you're missing a lot of assumptions in that statement. Try to be precise.

Force at right angles to the direction of travel changes momentum by changing direction, not by changing magnitude.

This is a frame dependent observation. The very same force on the same particle relative to a different inertial frame will change the magnitude of its momentum vector in that frame. Velocity and momentum are frame dependent, but angular velocity and momentum are not (sort of).

An orthogonal torque is applied, which exerts a variable force Ft with a max value as dm passes the north and south poles of its orbit and zero at the equator.

And I cannot make sense of this statement, which seems to imply that no force is exerted on an orbiting thing if it is over the equator at the time. I don't think you meant that, but I cannot figure out what you're actually trying to convey there. An orbiting thing doesn't pass its own poles. Those poles define its axis of orbital motion.

The particle dm is part of a rigid mass.

Then it's not in orbit. Not sure why you're bringing up orbital mechanics to describe the precession of a rigid object. The physics is different.

Yes, your description of the precessing path of a particle over a ball seems accurate enough, and is not unlike the path traced over Earth by a satellite with a polar orbit. But that satellite orbit isn't precession, it's just the Earth rotating underneath it in that case.

The equal and opposite momentum generated by the contra rotation of the gyroscopes is turned through 90 degrees in reaction to equal and opposite torque, resulting in the positive momentum at the red axis, the vector sum of the original momentum about the yellow axis.

Your diagram 2 (reinterpreted as momentum instead of velocity as labeled) doesn't do its vector addition correctly.  It seems to imply (via the dotted line) that all vectors have the same magnitude, but if I add OA to OB, I get OD except with √2 greater magnitude. This becomes more apparent if vectors OA and OB have significantly different magnitude, as is typical.

There is always the other way to operate the device.
In the OP the process started by contra-rotating the two gyroscopes about their yellow axis.
This time, do something different, start the motor attached to the red reaction flywheel which rotates the flywheel about the red axis and counter rotates the two gyroscopes, also about the red axis. The gyroscopes are both spinning in the same direction. The sum of their angular momentum is equal and opposite to that of the flywheel. This generates momentum in equal and opposite quantities. You could if you wanted to, detach the flywheel, take it away and use the momentum that has been generated to tighten a bolt.

Yes, this is the angular momentum of 'assembly A' of which I spoke. It includes the precession momentum if the gyros are running.  Precession does have momentum, and it needs to go somewhere when external torque is removed. It isn't often discussed in the typical youtube videos.

Uncouple the red flywheel. The two heavy gyroscopes and the light frame will continue to spin about the red axis. The flywheel will continue to spin.

Start the yellow gyroscope motor to counter rotate the rotors about the yellow axis.

This movement, the physical rotation by the drive belt of the spherical rotors about the yellow axis is orthogonal to the spin of the gyroscopes which is about the red axis.

The two gyroscope rotors are spinning about the red axis and therefore the orthogonal rotation about the yellow axis is, by definition precession.

Um, no.  The red axis is fixed, and the yellow axis is changing at the rate of rotation of the red axis. That makes the yellow axis the spin and the red the precession. It's all about which axis changes, and not which one spins up first.  So even if the yellow motor gives very low RPM to the yellow axis and the red one is spinning like crazy, that's just very fast precession of of a low-speed gyro. Yes, the angular momentum of the ball at any moment is nearly aligned with the red axis in this case.  If you suddenly severed one yellow ball from the system, the ball would be rotating along some axis that is close to but not quite aligned with red.

The dynamics of a gyroscope dictate that there is a torque present at the blue axis, again by definition. Each of the rotors will generate a torque. The torques will be equal and opposite.

Yes.

When a rotor spinning about an axis(red) is subjected to a torque about an orthogonal axis (blue) there is precession about the other orthogonal axis (yellow)

But red does not precess about the yellow axis in this case, so this is wrong. The assembly A, spinning with the red wheel detached, will continue to spin at an unaltered rate when the yellow motor spins up the two gyros. It will not precess at all about the yellow axis, only the red. Doing otherwise would violate conservation of angular momentum.

OE is the vector sum of the actual physical rotation of the rotor about the yellow axis and the actual in the real world physical rotation of the rotors about the red axis.

Except as I pointed out before, you did that addition incorrectly. The light dotted square you drew does not meet points B and A like it should. Look up vector addition on the web.

[Momentus (OP)]

I posted a drawing of a device that creates momentum. I thought that there was sufficient detail in the drawing for it to be replicated, as a physical or as a virtual machine.

I also thought that there would be some interest in the fact that it creates momentum, this is a science forum and the ramifications are profound. It should have been, could have been exciting.

I did try to explain how the device does work. Please note works. Did anyone try to follow my explanation?

@Halc It is evident from your posts that you do not know very much about gyro-dynamics.

It is not labeled, but I presume the reaction motor is that grey thing that spins up the red wheel. Also not labeled is whatever you consider the precession axis. There is the one main axle in the whole setup which is the rotation axis of the red wheel.  There is no precession axis

Spin axis Yellow
torque axis blue
precession axis red

There is a precession axis. The Red axis is the precession axis. Which is an essential part of the way the device works. You did not grasp that at the start of this thread, simple basic gyroscope theory and you do not understand it. Not then and not now.
I does seem that I am posting for your educational benefit, so much of your stuff is alternative facts. I would have enjoyed discussing Anomalous Dark Motion, not having to endlessly repeat basic gyroscope theory.

This makes a lot of assumptions, primarily that Fc is not continuous but actually changing over time.  A continuous force on a particle actually causes a more or less parabolic trajectory.
The statement makes the assumption that the particle is in the vicinity of point O.  That point could be on the other side of the galaxy, and a force Fc exerted at right angles to its motion (in frame F) is hardly going to put it in orbit about O.

Everything in this statement is wrong, part of your fantasy alternative facts world.

“This makes a lot of assumptions,” It is not my assumption “primarily that Fc is not continuous” It is your very own very strange idea that centripetal force (Fc) changes over time. And that “A continuous force on a particle actually causes a more or less parabolic trajectory.”

“on the other side of the galaxy, and a force Fc exerted at right angles to its motion (in frame F) is hardly going to put it in orbit about O.” I will bear this point in mind when I design a device which requires a force that spans the galaxy.

Um, no.  The red axis is fixed, and the yellow axis is changing at the rate of rotation of the red axis. That makes the yellow axis the spin and the red the precession. It's all about which axis changes, and not which one spins up first.  So even if the yellow motor gives very low RPM to the yellow axis and the red one is spinning like crazy, that's just very fast precession of of a low-speed gyro. Yes, the angular momentum of the ball at any moment is nearly aligned with the red axis in this case.  If you suddenly severed one yellow ball from the system, the ball would be rotating along some axis that is close to but not quite aligned with red.

This is just gabble and there is no one reading your rubbish who cares enough to correct you.

[Momentus (OP)]

Not much point in posting on this thread. However the subject is of such a fundamental nature that I feel I should try

The current state of American politics shows the power of belief The Dunning-Kruger Effect. I like to think that I don't do that. I do not “believe” the current paradigm, only that it gives theories that are the best explanations.

If it disagrees with experiment, it's wrong. In that simple statement is the key to science. Richard Feynman.

The three people who posted on this thread “believe” that momentum is conserved. I guess that is the same for everyone else who has read the thread.

The twin rotors device can be given the grand title of experiment. It works in a universe governed by Newton’s laws, it obeys those laws. Therefore it can be analysed by the laws.
As @Halc demonstrated with considerable effort, if you wish to prove that it does not work then you must also change Newtons laws. (Multiply, yes. It was adding apples to oranges that doesn’t work )

So to continue.

First a reprise of the basics.

A gyroscope’s Spin, torque and precession are orthogonal.
Torque results in orthogonal motion
Motion results in orthogonal torque.
Precession does not change the magnitude of angular momentum only the direction.
(Force must move mass along the line of action to do work).

Now on to my vector drawing.
There are two axes, red and yellow. The red axis could be said to represent, for example the flow of a river, the yellow axis could be the speed of a ferry boat. As the boat crosses the river, its actual path can be plotted.

It could represent a trackball, controlling the cursor of a computer. As the ball is moved it rotates sensors placed at right angles. If only the yellow sensor is rotated, a horizontal line is drawn.  If only the red sensor is rotated, a vertical line is drawn. To draw a curve, both are rotated.

I choose my vector drawing to represent the physical motion of the twin rotors experiment. The red axis represents the rotation, the angular velocity about the red axis, the yellow axis represents the rotation, the angular velocity about the yellow axis. No more than that and certainly no less.

So I can ask the question what happens if I rotate it first about the yellow axis with a magnitude of OA then rotate it about the red axis with a much smaller magnitude of AC?

Obviously the vector is changed in direction to OC. Less obvious is the fact that, in the real world model which the diagram represents, the magnitude of the yellow vector is reduced.

Furthermore I can ask the question what happens if I rotate it first about the red axis with a magnitude of OB then rotate it about the yellow axis with a much smaller magnitude of BE?

Obviously the vector is changed in direction to OE. Less obvious is the fact that, in the real world model which the diagram represents, the magnitude of the red vector is reduced.

This becomes clear when the magnitude of of both rotations is changed to give the resultant OD.

The changes in magnitude in the red and yellow vectors are governed by the fact that the resultant spin speed must remain constant. (precession does not change the magnitude).
The vector drawing shows that starting with rotation(spin) about either axis and then adding rotation about the orthogonal axis, (precessing) will move the resultant spin axis to the OD vector. Returning this rotation (precession) back to zero will return to the original vector condition. As one would expect.

The vector drawing also shows how the angular velocities must be manipulated to change the spin vector all the way from OA to OB and vice versa, whilst maintaining a constant magnitude of spin.

To conclude:
Equal and opposite forces are used to generate equal and opposite changes in momentum in accordance with Newton’s axioms.

The resultant vector sum gives a positive value. Momentum is created

That is the anomaly which I call Dark Motion.

It manifests in the Newtonian Universe as Dark Matter and as Dark Energy.

Build your own model, that is the only proof you will need.

[Momentus (OP)]

So where did I go wrong?

2. Run the reaction motor. This rotates the Gyroscopes about their precession axis, accompanied by equal and opposite torque orthogonal to the spin axis. There can be no torque exerted by the reaction motor along a precession axis, therefore no substantial movement of the reaction flywheel.

It is not labeled, but I presume the reaction motor is that grey thing that spins up the red wheel. Also not labeled is whatever you consider the precession axis. There is the one main axle in the whole setup which is the rotation axis of the red wheel.  There is no precession axis
since there's only the motor producing spin and counterspin. So the red wheel spins one way, and the contraption to the left spins the other way.  Still zero net angular momentum.

I assumed @Halc was being serious. He was actually taking the piss. Wind up the Newbie  and watch him struggle to explain.

I hope you all enjoyed the Newbie roasting.