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23 November 2024 15:19
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Question |
| Asked by: |
Blaze |
| Subject: |
Actual “Steady State” precession in a gravity powered overhung gyroscope doesn’t exist |
| Question: |
A “perfect” overhung gyroscope that is gravity powered will have the following characteristics:
1) Steady state precession can never be fully achieved.
2) It will continue to accelerate over time.
3) It will continue to drop slowly over time.
4) The actual cause of continued precession is the continued tilting of the flywheel. It is not caused simply by a force acting on the flywheel. (See the last paragraph for further explanation of this point)
A “perfect” gyroscope would be defined as one that has no friction of any kind, has no dead (non spinning) mass and has all mass distributed ideally, however it does have mass and therefore, inertia. This perfect gyroscope will continue to drop because the acceleration to steady state speed is asymptotic and therefore steady state speed can never be fully reached. I have done the experiments and have seen the curves that I filmed during the acceleration phase (commonly called the “Drop”). I have also mathematically done the integration in graphical format for the acceleration phase and it matches the experiments. The acceleration to steady state speed is asymptotic. This means that the gyroscope is always slightly accelerating because full steady state speed can never be reached.
Since steady state speed is never fully reached, the counter gravity couple is always smaller that the gravity couple, so the gyroscope continues to fall very slowly. I have, in other threads on this forum, talked of an “up force” which would more correctly be called an “up couple” or a counter gravity couple.
The concept of a counter gravity couple was also described by Glenn Hawkins (and maybe by others as well) elsewhere in this forum. Although Glenn didn’t call it a counter gravity couple, the description he provided was in fact, just that.
Glenn Hawkins: “A gyroscope does not act any differently whether the force to cause precession comes vertically from gravity, the fall, or comes horizontally from mechanical force, the push. Either way the particles in the flywheel are deflected to cause right angle torque, which causes the right angle movement. This is precession as you know and as we see it in an overhung, gravity powered demonstration. The precession movement itself causes a secondary right angle torque. That is, one twist causes yet another twist. This secondary torque caused by precession twists down of the pivotal area and holds the gyro aloft. If there is enough angular momentum in the wheel it will hold itself up, plus the axle and frame of the gyro as it slowly, slowly falls. The lift force can never be greater than the combined gravitational force that pressed down on all the parts and sets everything in motion. This is why gravity twisting back upon itself, can never cause a rise, but only a sustained slow, sometimes incredibly slow decent.” Glenn had it exactly correct and since the gyroscope is continually falling, it is also continually tilting.
So, steady state precession is never actually reached because acceleration is asymptotic. That means that the gyroscope is always accelerating which means that that the counter gravity couple (which is secondary precession or “yet another twist” as Glenn puts it) is always slightly smaller than the gravity couple. That means that the gyroscope is always dropping and always tilting. All of which means that in the real world, the reason the gyro continues to precess is because it is continuing to tilt because it is continuing to fall.
It is important to note here that continued precession may happen without continued flywheel tilting in a “perfect” gyroscope IF steady state speed could be fully achieved. Full steady state speed would provide a perfect balance between the gravity couple and the counter gravity couple and without any friction in the “perfect” gyroscope to slow things down, the momentum achieved by the flywheel during the acceleration phase would cause it to continue coasting without any further tilting of the flywheel required. However in a real world gyroscope, even if full steady state speed could be reached, the gyro would continue to fall due to all the frictions, dead weight and imperfections involved that work to slow it down to something less than steady state speed.
As usual, thoughts and comments are welcome.
Blaze
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| Date: |
11 December 2012
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Answers (Ordered by Date)
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| Answer: |
Nitro - 11/12/2012 19:02:17
| | | Good evening Blaze,
I think Nitro’s law bears repeating:- “A gyro will precess every force applied to change its axial angle – including those forces you have overlooked”.
A drop does not cause precession, rather, precession of the friction of the vertical pivot causes downward precession which is the drop.
Kind regards
NM
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Blaze - 12/12/2012 01:13:19
| | | Hi Nitro. True, the "drop" doesn't cause precession. It is the tilting of the flywheel that causes precession and when starting the gyro this is observed as a drop in vertical height.
Blaze
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Nitro - 12/12/2012 08:11:00
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Good morning Blaze,
Sorry Blaze, I thought you were referring to the slow, continuous, droop of the axial angle caused by the pivot friction being precessed. The momentary starting drop in the axial angle is something else, but I think the Nito’s first law explains this as well.
“A gyro will precess every force applied to change its axial angle – including those forces you have overlooked”. Let us look more closely at the start point:- the gyro is released and gravity acts to try and change the axial angle so the gyro precesses at ninety degrees to the force, in this case gravity, acting to change its axial angle. Woops! We have overlooked the inertia of the gyro’s non spinning - or insufficiently spinning - component under that moment of initial, almost instantaneous acceleration, into precession. What direction will that momentary inertia be precessed into?
Kind regards
NM
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Glenn Hawkins - 13/12/2012 06:04:51
| | | Outstanding, Blaze. That is a very fine explanation. My complements and it is pleasing to follow this thinking. I will now begin to think of these mechanical laws when grouped together as coined the condition, ‘study state precession’. We tip our cup to you all in happiness on New Years Eve.
Cheers,
Glenn,
(I too am working again, a little bit. How I hope I continue to actually finish something.)
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Glenn Hawkins - 13/12/2012 06:33:10
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If the circumference and mass of the flywheel is greatly and contentiously increased; while a mechanical force is applied cause tilt down is greatly, but equally increased, what would be the result?
(ONE) Precession would not increase in speed?
(TWO) Precession would increase in speed?
(THREE) Maintained speed, but collision force would be greater?
(FOUR) Momentum and inertia would not increase?
Glenn,
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Glenn Hawkins - 13/12/2012 06:36:48
| | | (FIVE) The speed of tilting (dropping) would remain unchanged?
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Blaze - 15/12/2012 03:00:34
| | | As Sandy Kidd once said, a gravity powered system is a system in decay. Of course it would be relatively easy to correct the 4 factors listed at the beginning of this thread by simply supplying some rotational energy to the pivot.
regards,
Blaze
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Blaze - 23/01/2013 02:02:57
| | | Further reflection on this concept has brought forth new insight. “Steady state” precession would NOT be possible if FULL steady state speed could be achieved. As a matter of fact, precession would cease.
The only reason a gyro precesses is because the gyro is being tilted. This happens continuously with a gravity powered gyro because, as stated before, full steady state speed cannot be reached. Therefore there is always some tilting of the gyro happening, even if it is a very small amount. When all tilting force (gravity couple) is removed from the gyro, it stops precessing. Anyone working with gyros for any length of time knows this and has seen it repeatedly. So if full steady state speed is reached, the counter gravity couple would exactly equal the gravity couple and the gyro would not have any tilting force applied and it would stop precessing.
In the past I have stated that the reason the pivot end of the shaft stay on the pivot during slow “steady state” precession is because the two different motions, precession (turning about a vertical axis through the flywheel center of mass) and momentum (straight line, like in the tree house experiment), are perfectly balanced which cause a natural curving motion that is centered on the pivot. If the gravity couple is suddenly removed or perfectly counter balanced, then the precession motion ceases and so does the balance of motions it was a part of. However, this would not stop the momentum. It would still exist, which means that the gyro would want to continue in a straight line. This momentum would pull the gyro off the pivot or pull the pivot along with it.
Best to all,
Blaze
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Nitro - 23/01/2013 11:47:10
| | | Hi Blaze
Well observed, as always. Your description as to why a gyro displays no rearward torque on its pivot is good and, I believe, sound. However I don’t believe that the axial angle needs drop to “drive” precession.
Only a torque need be applied, as if to change the axial angle, to cause precession. This applied torque, amazingly, does not have to actually move the axle at all to “cause” precession. Any axial drop in the vertical plain is caused solely by friction in the horizontal plain resisting the gyro’s gravity applied torque caused precessional movement, being itself precessed (remember Nitro’s first law) downwards in the vertical plain. Sorry about the overlong sentence – you can breath again now!) Reduce the friction and you reduce the axial drop – if only we could do away with friction altogether!
But that would bring us into the realms of over-unity lunacy and I think we have enough on our plates getting round the second an third laws to keep us busy for now.
Kind regards
NM
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Luis Gonzalez - 25/01/2013 01:53:12
| | | Hi Blaze,
Thanks for the Great above postings of 23/01/13 and 11/12/12.
Your argument is well presented in a compelling manner. Unfortunately it doesn’t prove this theory is better than the previous one that claimed steady-state orbit coasts (against minor Friction forces).
In your posting of 11/12/12 you say that a “Perfect Gyro” orbits “always-accelerating” asymptotically, and this acceleration is tied to the gyro’s downward motion (assumedly because steady-state orbiting-precession doesn’t coast but instead requires some small source of energy from gravity).
Am I paraphrasing correctly?
My thoughts on this are as follows:
Since the resulting orbiting acceleration is asymptotic, then the required downward driving-motion will also need to be asymptotic.
As a rule, asymptotic motions fade quickly and become undetectable but that is not quite what we observe. Observation tells us that the downward motion actually accelerates, and this is more consistent with the effects of “Frictions”.
Regarding concurrence of math with observed experiments here are my thoughts:
Integrations into math from experimental-results always reflect the imperfections of the experiment (it can cause circular logic where the math and experiments support each other).
(Also, am I interpreting wrong, or did you actually claim that “Perfect” gyros never achieve steady-state (at the beginning), and that they can in fact achieve steady-state (at the end of first posting)?
Regarding your posting of 23/01/13, the fact that removing the torque makes the gyro fall is consistent within your theory; however it is also consistent with the alternate theory that claims steady-state orbit coasts against minor Frictions. So we don’t have conclusive proof either way.
Perhaps we can approximate which theory is more accurate by determining friction’s magnitudes and comparing them to:
- The force necessary to hold in place a gyro against centrifugal force of 1/12 Newton (which is about 0.0187341 Pounds of force, or 0.6008 lb-ft/sec-sec)
- And the force necessary to cause the downward motion of orbiting precession.
Regards,
LuisG
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Blaze - 25/01/2013 02:09:57
| | | Nitro:
"Only a torque need be applied, as if to change the axial angle, to cause precession. This applied torque, amazingly, does not have to actually move the axle at all to “cause” precession."
I used to think the same thing until I realized that the total absence of torque or the perfect balance of opposing torques does the same thing. Neither situation will cause the gyro to precess.
I did a couple of physical experiments and a thought experiment to prove that some tilting (and therefore movement) is required, even if it is just a very small amount.
The thought experiment is this. If you had a gyro flywheel spinning in space, free of any gravitation field, it would not precess. That is the total absence of torque.
The physical experiments are these. I took a Tedco gyro and had it precessing on the pivot. I was holding the pivot over my head while the gyro was precessing. I pulled down on the pivot extremely quickly so that the pivot was moving down faster than the gyro was falling. As soon the pivot moved away from the gyro axle the gyro started to free fall and stopped precessing. One could argue that is the perfect balance of torques (same torque on both sides of the gyro axle) or the total absence of torque (because it is in free fall).
The other physical experiment is having a gyro spun up and on the pivot but also having a support under the outer end of the gyro axle (side not on the pivot) and then applying force straight down on the axle that has the support under it. Of course the gyro does not precess. It wouldn't matter if the support had zero friction on the surface it was resting on, the gyro still would not precess. That is the perfect balance of opposing torques (forces).
All of this does not prove that the gyro has to tilt to cause precession but rather it proves that if there is no tilting the gyro does not precess.
So if a gravity powered gyro were actually precessing at "full" steady state speed (not actually possible of course), there would be a perfect balance of the gravity couple and the counter gravity couple and the gyro would not precess. However the orbit speed it has (momentum) would still be there.
regards,
Blaze
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Blaze - 25/01/2013 03:06:29
| | | Hi Luis.
Luis: "In your posting of 11/12/12 you say that a “Perfect Gyro” orbits “always-accelerating” asymptotically, and this acceleration is tied to the gyro’s downward motion (assumedly because steady-state orbiting-precession doesn’t coast but instead requires some small source of energy from gravity)."
Blaze: You are paraphrasing mostly correctly. Steady state orbiting precession is not reached instantaneously as you know. It takes some small amount of time to get up to speed. As the gyro is falling it is accelerating towards steady state. However, the full steady state orbiting precession can never be reached so the counter gravity couple never fully equals the gravity couple, so the gyro continues to fall, at an ever decreasing rate.
Luis: "My thoughts on this are as follows:
Since the resulting orbiting acceleration is asymptotic, then the required downward driving-motion will also need to be asymptotic.
As a rule, asymptotic motions fade quickly and become undetectable but that is not quite what we observe. Observation tells us that the downward motion actually accelerates, and this is more consistent with the effects of “Frictions”."
Blaze: You are mixing up the "perfect" gyro as defined in the posting of Dec 11, 2012 with a real world gyro. The "perfect" gyro as defined in this thread does not have any friction (no pivot friction, no flywheel friction which means the flywheel doesn't slow down, no friction of any kind). So the continued fall of the "perfect" gyro would be exceedingly slow and almost undetectable. The real world gyro does have frictions of various kinds so the fall is far more rapid. Added to the problem is that in a real world gyro, the flywheel is slowing down which means the gyro orbiting-precession has to speed up. The only way it can get energy to speed up is through additional potential energy so the gyro falls more quickly as the flywheel slows down.
Luis: "Regarding concurrence of math with observed experiments here are my thoughts:
Integrations into math from experimental-results always reflect the imperfections of the experiment (it can cause circular logic where the math and experiments support each other)."
Blaze: I did not "integrate into math form experimental results". I used what I have learned/discovered about what is happening during the acceleration phase to graph the acceleration's motion, which is an asymptotic curve. I did not use the experiments in any way to do the math. The math was done independently from first principles and the graphical result was compared to the experiments. The two curves match very nicely.
Luis: "(Also, am I interpreting wrong, or did you actually claim that “Perfect” gyros never achieve steady-state (at the beginning), and that they can in fact achieve steady-state (at the end of first posting)?"
Blaze: I said in the end of the first posting that IF (note the IF) a gyro could achieve full steady state speed (which is not possible of course under gravity power alone), then coasting would take place and continued tilting would not be required. This was revised in the posting of Jan 13, 2013 and a further explanation given in the posting of Jan 25, 2013 to Nitro.
Luis: "Regarding your posting of 23/01/13, the fact that removing the torque makes the gyro fall is consistent within your theory; however it is also consistent with the alternate theory that claims steady-state orbit coasts against minor Frictions. So we don’t have conclusive proof either way."
Blaze: This thread is about what would happen with a "perfect" gyro as it was defined in the thread and that definition was that it had no friction of any kind. So any continued falling of the "perfect" gyro would not be from friction because by definition it doesn't have any. Of course in a real world gyro, any friction will exacerbate the fall of the real world gyro.
regards,
Blaze
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Nitro - 25/01/2013 16:53:39
| | | Dear Blaze,
Both your practical and thought experiments are flawed as they both remove the very torque we were torqueing about. It won’t precess without a torque. A better way for you to grasp this concept might be to look at it from a different angle (here in the British Isles we have to contend with insane European metric angles – what is an Angular Merkel, anyway).
Instead of trying to do without torque try instead increasing and decreasing the friction on the pivot. With your excellent grasp of things mathematical you should be able to project different frictions/drops to the (impossible to achieve) imaginary point where friction disappears. Your projection to “a no friction state” will show that the gyro no longer drops at all (because the drop is solely caused by pivot friction being precessed into the downward drop) and yet it will still precess because it has a torque (not drop) applied to it.
Enough of this torque talk, it’s Burns night and my birthday so I’m off to shake the moths out of my Erach Cameron kilt, stuff some bagpipe proof earplugs in me lug ‘oles and sink a Glenmorangie single malt to brace myself for tonight’s bash. Awah the noo, ye wee sleakit timrous beasties. Lang may yer lum reak! An aw tha kind o stuff.
Kind regards
NM
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Blaze - 26/01/2013 04:20:16
| | | Nitro:
"It won’t precess without a torque"
Blaze:
You hit the nail right on the head. It won't precess without a torque. That is the whole point. At "full steady state" speed the counter gravity couple exactly equals the gravity couple and there is no torque acting on the gyro so the gyro won't precess. Or rather, to put it more accurately, since it is already precessing before it reaches full steady state speed, it would stop precessing when it reaches full steady state speed.
Cheers,
Blaze
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Harry K. - 26/01/2013 13:14:48
| | | Hello Blaze,
Unfortunately Nitro is right. The "counter gravity couple" is caused by the 90 degrees deflected torque in precession plane. Friction and other and other losses, e.g. energy losses to accelerate the mass of the gyro to precession speed, friction torque at the pivot, etc., will cause again a counter torque to precession torque which will cause in return the drop of the gyro.
If you would compensate these losses in form of additional torque in precession plane, which is equal to the torque of all losses, the gyro would not drop but endless precess. However, this additional torque requests the continuous input of energy.
Harry
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Luis Gonzalez - 27/01/2013 04:49:17
| | | Hi Blaze,
You sound very sure that one of your equations (which reflect minute changes of perfect gyros) matches experimental results (which reflect relatively large changes in the real world and includes frictions etc.)
What does your asymptotic equation represent, orbit velocity, or orbit acceleration, or something else?
Regards,
LuisG
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Blaze - 27/01/2013 17:43:37
| | | Hi Luis.
As you already know, the experiments that I did and recorded with a laser pointer attached to the axle of the flywheel, shows the path that the flywheel takes during the acceleration phase (commonly called the Drop). This path is due to the changing magnitude of the force couples that the flywheel experiences during acceleration towards steady state orbiting precession. If anyone wants to review what I mean by that you can look at my May 20, 2012 posting - http://www.gyroscopes.org/forum/questions.asp?id=1441 - where I went through the changing magnitude of the force couples during the acceleration phase. Although the math part of it is not 100% accurate (as it was based on an incorrect linear acceleration rather than the correct asymptotic acceleration), the description of the changing magnitudes is actually VERY close to what is really happening. You should also replace the word "force" with the word "couple" when reading that posting.
The graph I did from first principles of the acceleration phase was done to about 97% of steady state. When I said that the graph matched the experiment curves very nicely, it did not mean that they matched perfectly, but rather were a relatively close match. There were differences which I believe are due to the real world friction that the real world gyro experiences. The calculated acceleration time and the experiment's acceleration time do differ with the calculated time being somewhat shorter. This is because the frictions of a real world gyro will slow the acceleration to steady state and also increase the vertical drop distance. In other words it takes longer to get to steady state so the curve from the experiment and the curve from the calculation will be close but somewhat different.
regards,
Blaze
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Luis Gonzalez - 28/01/2013 02:02:10
| | | Hi Blaze,
Orbiting precession would not be the first time that “static-forces” (forces that do not advance forward) create interesting motions.
Some prime examples include satellites, and different-shaped spinning objects, which have a centripetal component that does not advance forward while these force sustains interesting angular motion. This is very important.
Harry made an excellent point when he alluded to adding a torque in the precession plane, i.e. a torque that hurries precession’s rate (I know this is a concept familiar to you Blaze, as we have previously spoken about it).
Harry’s interesting point demonstrates that it is possible to achieve the “balance” where steady state orbit is achieved by compensating for frictions (and all other potential spoilers), and it is done without bringing any ill effects to precession…
I believe this proves steady-state can be achieved without knocking-down a gyro………
Having said that, I don’t want to throw out the possibilities you have presented; either theory can be correct.
A good theory needs to address all gaps directly.
Most important:
There is no convincing argument that the equation for precession’s acceleration is asymptotic to T / L, in respect to time.
Showing proof of this quality will be sufficient for a couple of us in this forum.
Experimental proof is probably not possible without eliminating all the frictions and relevant deadweight.
Cheers,
LuisG
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Blaze - 31/01/2013 00:28:49
| | | Luis, The acceleration is asymptotic. I will contact you privately on this matter.
regards,
Blaze
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