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| Asked by: |
Blaze |
| Subject: |
Gyro precession and motion in space |
| Question: |
The gyro is in a non gravitational field and we will ignore all frictions for this question.
Imagine a gyro system that consists of a spinning flywheel at one end of the axle and a non spinning wheel of exactly the same dimensions and mass welded to the other end of the axle. If neither wheel were spinning and you put some torque on the axle perpendicular to the length of the axle and then just let the system go with no further input, the system would simply barycenter in one plain of rotation (let's say horizontal plain for this discussion).
Now, if the flywheel were spinning at high speed and you put some torque on the axle perpendicular to the length of the axle and then just let the system go with no further input, what would the resulting motion be?
cheers,
Blaze
P.S. In both scenarios the application of torque would be with a force couple at the ends of the axle equidistant from the barycenter (or the torque could be applied at the barycenter point) so that no linear movement would occur during precession/rotation of the system. |
| Date: |
2 April 2013
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Answers (Ordered by Date)
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| Answer: |
Harry K. - 02/04/2013 23:52:19
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| | Hello Blaze,
I really cannot imagine that you do not know the answers.
Or do I miss something in this matter?
Regards,
Harry
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| Answer: |
Blaze - 03/04/2013 03:10:07
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| | Hi Harry. I believe I do know the answer but I want to see what others come up with on the chance that I am perhaps over analyzing and coming up with the wrong result. I also believe there may be more happening here than this simple gyro system would lead one to believe. So if you don't mind, could you tell me what you think the resulting motion will be.
thanks in advance,
Blaze
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| Answer: |
Harry K. - 03/04/2013 22:34:57
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| | Hello Blaze,
I would say the applied torque causes the flywheel system to rotate (precess) in perpendicular direction around the centre of dead weight mass, i.e. around the bary centre, as long as the torque will be applied. The bary centre is the point of lowest resistance related to angular momentum in precession plane, although the precession torque will be generated around the centre of the flywheel's spinning mass perpendicular to the plane of the applied torque.
Regards,
Harry
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| Answer: |
Sandy Kidd - 03/04/2013 23:19:23
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| | Hello Blaze
Had a few minutes to burn so for what it is worth .
The rotating (mass) gyroscope will appear to lose weight as soon as radial acceleration takes place,
This will cause the non-rotating mass to descend and will by default displace the rotating mass (gyroscope) upwards until the shaft is vertical, or until the rotating (mass) gyroscope is running with its plane of rotation horizontal, which is the point of lowest action.
Regards,
Sandy.
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| Answer: |
Blaze - 04/04/2013 02:45:09
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| | Gentlemen, thank you for your responses.
Harry, I agree with what you are saying as long as the torque is applied but I am looking for what happens after the system is precessing and the torque is removed and the system is free to do whatever it will do. I don't believe you answered that, unless I completely misunderstood you.
I will reveal my thoughts on the situation after I see your answer.
cheers,
Blaze
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| Answer: |
Harry K. - 04/04/2013 21:21:13
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| | Hello Blaze,
During acceleration to precession velocity the gyro system would pivot around the centre of mass of the gyro system (bary centre) in the plane of the applied torque in addition to the perpendicular rotation in precession plane.
As soon as the applied torque will be removed, the stored angular momentum in precession will cause the gyro system to pivot back around the centre of mass in the bary centre to the initial position in the plane of formerly applied torque until angular momentum in precession motion has been transformed completely.
In this consideration I assume the applied torque plane will move together with precession motion so that the applied torque will be always perpendicular to the actual precession position.
In summary an applied torque wil cause the gyro system to precess around the barycentre, which is located in the centre of mass of the gyro system. After removal of the applied torque, precession will stop abd the gyro system remains at its actual position in space. There should be no difference in basic behaviour whether 1 or 2 flywheels are spinning at the same time.
Blaze, Your question was not so simple as I've thought at first sight.
I'm curious about your thinking regarding this issue.
Regards,
Harry
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| Answer: |
Blaze - 05/04/2013 02:31:54
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| | Hi guys. The question is far from simple and I believe it generates an interesting motion which I plan to test this weekend. But before I reveal anymore, I have a question or two for you to chew on.
Why would the system stop? In other words, why would the rotation of the axle about the barycenter stop? It would take a force equal to the initial input force to cause it to stop and I don't see where that force would come from because once the system starts moving the torque is removed and the system is not touched anymore, so why would it stop? When precession is happening there is angular momentum stored in the system. If the system stops rotating about the barycenter, where does the momentum go?
cheers,
Blaze
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| Answer: |
Harry K. - 05/04/2013 18:14:04
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| | Hi Blaze,
The rotation stops because there is no torque acting to cause a change of the spinning vectors of the flywheel's mass. As described in my last post the stored angular momentum in precession motion will be deflected back about 90 degrees into the plane of formerly applied torque and causes in return a rotation in this plane until angular momentum in precession is completly consumed or in better words transformed.
Precession motion only occurs as long as a tilting or pivoting torque is acting. After removal of this torque the flywheel is only spinning around its centre of spinning mass and remains at its actual position.
At this stage the spinning flywheel is a part of the basic reference system in the universe.
I'm curious regarding your experiment.
Regards,
Harry
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Blaze - 05/04/2013 19:34:38
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| | Harry: The rotation stops because there is no torque acting...
Blaze: Once the system is moving there is always torque acting. It is the force of motion of the system mass, momentum (mv). The motion of precession in the horizontal plain is "converted" to motion in the vertical plain. The original magnitude of the motion still exists (but in vertical plain). It is not consumed because the gyro doesn't have to climb up against gravity like it does on earth. Because the momentum does not get used up changing kinetic energy into potential energy, it still exists with the same magnitude, but in a different direction. The motion that is now in the vertical plain is providing the torque due to system momentum (mv). This "vertical torque" from mv will also change direction to the horizontal plain but in the opposite direction, which will change its direction to the vertical plain (downwards), which will change its direction to the horizontal plain, which will...... There is no "consumption" of momentum (magnitude remains constan), just a constantly changing direction due toi the gyroscopic effect of the spinning flywheel.
In actuallity, this change of direction will happen on a continual basis so the path of the two weights (spinning and non spinning) would describe circles that are the bases of cones with the tips of the cones at the barycenter.
The idea that the momentum doesn't cause continued movement of the mass sounds a lot like the arguments put forth by some people before the treehouse experiment I did last year.
cheers,
Blaze
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| Answer: |
Harry K. - 06/04/2013 09:05:52
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| | Hello Blaze,
You are right! In an environment of absence of gravity and friction obstacles the stored angular momentum (mw) in horizontal precession plane will be converted back into angular momentum in vertical plane and reverse.
However, I don't think the converting procedure will not happen in a cycling way between the perpendicular planes but rather there will be an equilibrium of angular momentum and motion in both planes, i.e. the gyro system will continously rotate in both planes at the same time around its centre of mass in the barycenter. The angular momentum is stored in both planes in equal parts.
Very good logical thinking work, Blaze. Well done!
Regards,
Harry
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Blaze - 07/04/2013 03:53:15
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| | I just performed the experiment and it worked beautifully. It did exactly what I thought it would.
I used two "precision tedco" gyroscopes, one spinning, one not spinning. The gyros were separated with mecanno straps by about 15.5 inches between the flywheel centers. I simply used a string though the center hole of the straps to hold the barycenter point and pushed down on the spinning side. The system did exactly what I thought. The two ends move in circles which are the bases of cones with the tips of the cones at the barycenter. The size of the circles depends on the amount of starting tilt but generally is between 2 and 4 inches in diameter for this system. The whole system does tend to drift a bit (slowly rotate in the horizontal plain). However, the drift very slow so I would say it is likely due to the imperfectness of the system (mecanno, string, rubber bands, no balancing of the components,etc.).
This would also work equally well if both gyros were spinning but the spins would have to be in the opposite directions for it to work or the motions would cancel each other out.
I plan on videoing the experiment but haven't got to that just yet.
blazing new trails,
Blaze
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| Answer: |
Blaze - 07/04/2013 16:27:12
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| | A few more experiments with the apparatus this morning. I usually provide the momentary input torque vertically.
1) Circles can be quite large, as much as 8+ inches if I provide a lot of input tilt.
2) Horizontal input torque works equally well, as expected.
3) Orientation of the system when the momentary input torque is removed is "one edge" of the circling motion, this too was expected because of the way the forces act during precession.
4) With both gyros spinning the circling motion is much quicker.
5) There is noticeably more gyroscopic resistance to the input torque with both gyros spinning.
6) Drift (rotation of the system in the horizontal plain) only seems to occur when the gyros are spinning slowly
cheers,
Blaze
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| Answer: |
Blaze - 10/04/2013 04:28:33
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| | Tried uploading a video of the experiment but our old digital camera uses some obscure format that wasn't recognized so I will have to try with a different camera.
cheers,
Blaze
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| Answer: |
Glenn Hawkins - 07/05/2013 12:30:53
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| | Harry was right at first before he changed. The little apparatus has its own built-in set of breaks. It will apply reverse force against all torques and movements when the applied force is removed. It cannot coast, but will stop itself. The only short duration of continued motion would be caused by dead weight, but that too would be overcome and stopped by the reversing breaks. Otherwise in space, while force was continuously applied as stated, the gyro would tumble in a chaotic jumble of motions that never could exactly repeat themselves.
I as well as you, have spent years reasoning out these things, and for me these breaks included and how and why they work. Now I am attempting to explain everything in a paper. I call it, ‘The Bible of Gyroscopes'. It was incredibly hard to reason and is incredibly hard to write and explain in a way that can easily be understood. If fact it is imposable. If I do my very best to simply, it will be difficult to understand.
Until I am finished, believe and accept, as you choose. We could never be lead to do otherwise anyway, for if we are if nothing else here, free thinkers. I have not furnished the paper.
I’m glad to be part of you.
Cheers,
Glenn
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Glenn Hawkins - 07/05/2013 15:49:21
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| | I should have added, the breaks play a critical roll in causing force to torque into a right angle direction.
Regards Glenn,
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Blaze - 08/05/2013 01:29:56
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| | Glenn "The only short duration of continued motion would be caused by dead weight"
The dead weight only wants to move in one direction (unless acted upon by a force). Only the action of the spinning gyro causes it to do otherwise. When I spin up both gyros, the apparatus continues for quite some time (very little dead weight that way). Only friction stops eventually stops the motion. There is no reason for the motion to stop if there were no friction and no change in flywheel speed. The momentum cannot be stopped without the same amount of force it took to get it started.
The motion is simply changed to move 90 degrees to the input force (momentum). Of course, as soon as it starts changing direction this becomes a new momentum input which the gyro again changes to 90 degrees, etc. etc. etc. Eventually it makes a complete circle. If you turn the wheels of your car and drive you will also make a complete circle because you are constantly applying a new input to the direction of motion.
Perhaps the problem here is that no one else seems to understand what is really happening during acceleration and deceleration of the gyro to "steady state" speeds which is a crucial part to understanding this motion.
It may be easier to understand this as simply precession in a different plain than one is used to because that is exactly what it is. If you could suddenly "remove" gravity from a precessing overhung gyro it would do exactly the same thing. Glenn, you should know why. It is because the counter gravity couple is still active, just the gravity couple has been removed.
Blaze
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Glenn Hawkins - 08/05/2013 03:39:22
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| | Dear Blaze,
You are so intelligent and I admire that particular commodity so much in a man that I actually wish I did not need to tell you that you are wrong. The apparatus could not work at all, without these built-in breaks that force back against continuing motion. No one has every known before how the gyroscope works. No, they never have! Not ever. Please note that I said, “Until I have finished my paper, believe and accept as you choose. We could never be lead to do otherwise anyway, for we are if nothing else here, free thinkers.
Bless you. I am proud of you mind,
Glenn
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Glenn Hawkins - 08/05/2013 16:32:07
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| | Hi Blaze,
You could be right. There are the breaks for certain, but like a spring, they compress then kick back. As we’ve always known, there are two conditions that support the gyro from falling. One condition goes like this; downward force (1) causes precession (2) causes lift (3) lift as it rises is stopped by the downward force (1) which began it all. Remove the gravity and your question is troublesome.
Still you could be wrong. I am not hurrying to finish this paper to publish. It is too important. I am glad you called this into question. Please try to confirm you hypothesis and I will try also. I don’t know where to begin as yet.
Good show,
Glenn
P.S. I understand dead weight.
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Blaze - 09/05/2013 02:16:47
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| | Hi Glenn. I already confirmed this with a simple experiment using a matching pair of tedco gyros spaced 15 inches apart with a string holding it up at the balance point (barycenter). With a momentary input (either horizontally or vertically) the motion is as I described. I also did the experiment with a small extra weight on one side to get the apparatus precessing and then removed the weight and again, the motion was as I described.
cheers,
Blaze
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Glenn Hawkins - 10/05/2013 12:28:02
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| | THANK YOU ALLOT : - )
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| Answer: |
Blaze - 04/03/2015 03:49:49
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| | https://vimeo.com/121209507
Two identical gyros (except for colour) separated by 15 inches with a string through the balance point to hold the apparatus.
A small momentary input force, either in the horizontal or vertical direction causes the gyro to "precess" in the vertical plain. The reason the apparatus comes to a stop is simply due to friction which is probably fairly high from the string rubbing on the metal straps as the apparatus precesses in the vertical plain.
Although it is not shown on this video, when a small weight is added on one side, the apparatus will, of course, precess in the horizontal plain. When the weight is suddenly removed the apparatus "precesses" in the vertical plain.
Blaze
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