|
Main Forum Page
|
The Gyroscope Forum |
3 May 2024 12:31
|
|
Welcome to the gyroscope forum. If you have a question about gyroscopes in general,
want to know how they work, or what they can be used for then you can leave your question here for others to answer.
You may also be able to help others by answering some of the questions on the site.
|
Question |
| Asked by: |
Nitro MacMad |
| Subject: |
precessional cause |
| Question: |
Dear friends and shed dwellers,
Apart from the input require to pre-spin and position a gyroscope, its precession requires no input of energy; only the application of a force. That we are used to saying “it takes energy to move a mass” should not be allowed to blind us to the fact that as far as a gyro is concerned its precession is its equivalent of being stationary. The only energy absorbed by a gyroscopic assembly is that from bearing, pivot and surface air friction. No energy is absorbed by, or require for, its precession.
To stop a precessing gyro only requires the application of a force equal (but of course not opposite) to the force causing its precession . In the case of the classic terrestrial toy gyro precessing around its tower because of the application of gravity to its mass, the force required to stop its precession is the exact equivalence of its stationary weight (mass under gravity) measured at the tip opposite its pivot.
Kind regards
NM
|
| Date: |
6 November 2007
|
| report abuse
|
|
Answers (Ordered by Date)
|
| Answer: |
Glenn Hawkins - 07/11/2007 17:36:40
| | | Dear friend and shed dweller,
A big yes to you for the complete post and a bigger hurrah pacifically for the statement below.
“…the force required to stop its precession is the exact equivalence of its stationary weight (mass under gravity) measured at the tip opposite its pivot.”
Exactly, in all details! I was really happy to learn this. It’s beautifully logical.
My curiosity forces a question. How are you using ‘terrestrial’? I know perfectly well you have a reason and an answer and knowing your mind a little bit it’s probably going to be wild.
You state, “…as far as a gyro is concerned its precession is its equivalent of being stationary.”
This is a special can of worms for me. It goes back to the general belief here that there would be no momentum, or centrifuge from precession when using a perfect gyroscope. In my mind I have arguments there has to be kinetic energy, potential force if you prefer, stored whenever mater is in motion. The potential remains until you attempt to use it, then all hell breaks lose ‘sometimes’. No matter though how you attempt, nor whether the chaotic reaction is easy and smooth, or wild you get nothing useful from what was, but is no more. The potential vanishes upon collision. However I further reason and have some strong test indications that there is two possible ways to use this hidden disappearing force, or kinetic energy, potential force, or even the concept of power depending on how are thinking in what terms. (Yes, I understand.) If I am wrong about this hidden potential then more than 99% of all apparatus’ intended to be built will fail. How sad. Anyway, bravo to you. I’ve learned something new. Thank you, Momentus.
Regards,
Glenn
|
| Report Abuse |
| Answer: |
Glenn Hawkins - 07/11/2007 20:48:45
| | | Dog-gone! The half quote I made of Momentus’ statement, “…the force required to stop its precession…” is a bit misleading without the whole quote. The force to stop is a right angle vector to precession he as much as said— it is vertical. Just read his entire last paragraph and all will be right. Maybe I am actually getting punch drunk. Excuse I.
|
| Report Abuse |
| Answer: |
Nitro MacMad - 11/11/2007 17:55:41
| | | Dear friends and shed dwellers,
THE STRANGE EFFECTS OF PRECESSION
There has long been debate here as to whether a gyro effectively (please note the word effectively) looses mass in precession.
Though some effects of normal (Newtonian) mass rotation can be observed in the classic terrestrial gyro on an Eiffel tower (sorry to disappoint, Glen, but the “terrestrial” has no “wild” explanation here, it’s just to show that the gyro is somewhere that is able to have gravity act upon it) these effects are much less than would be expected for the gyro’s mass and can be reasonably explained as being caused by the non gyrodynamic mass of the gyro’s cage.
Perhaps an easier demonstration of the effective loss of mass in precession is to fast frame photograph the acceleration from standstill to full precessional speed of a released gyro – it is impossibly fast! And I mean really impossible. There is no way for any (Newtonian) mass to be accelerated under gravity as fast as a heavy gyro released into precession. It is as near instant as makes no difference.
Oh! Incidentally, the reason that a gyro drops slightly (generally; the slower its rotation in relation to its mass/diameter the greater the drop), then largely recovers, is not because it is somehow sucking energy out of gravity to power its precession but because of the hefty instantaneous bearing load drag being precessed (remember Nitro’s first law) at the moment of release because of its near infinite acceleration.
Kind regards
NM
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 16/12/2007 14:49:10
| | | Dear Nitro,
Your brief statements speak volumes (visualizing precession as the stationary equivalent for a spinning gyro has provided food for thought).
Whether precession involves energy exchanges or not, is partially obscured in a number of ways.
First there is the intended meaning of “using energy” Vs “consuming energy”; if there is a difference between “common conversions” of energy and “absorbing” energy, I suppose the difference is in the level of entropy produced by the conversion.
Second there are different segments in the onset of precession; are we talking about precession’s steady velocity, or its acceleration from zero to a steady angular velocity?
Finally, is steady precession a monolithic motion, or does it consists of multiple seamless events/motions?
I am going to assume we are referring to energy exchanges rather than high entropy conversions (such as friction). Therefore I must ask, at what point in the following list of interactions do energy exchanges/conversions cease to occur?
Is there use of energy when:
1) Lowering or raising a mass of 1Kilo by 1 meter
2) Inducing velocity to a mass of 1Kilo
3) Changing the direction in the existing motion of the 1Kilo mass
4) Starting a still mass of 1Kilo into a spinning motion
5) Changing the orientation in the spin-motion of the 1Kilo spinning mass
Which of these are and which are not representations of an energy exchange?
Why and/or why not?
Consider the following:
Precession is perceived as a single fluid motion but may be more complex than it appears on the surface. We can alternatively perceive precession as a compound motion, composed of two main sub-components of motion; first “reorientation” and second “displacement” or translation.
Precession’s reorientation motion can occur without “displacement” or translation (as illustrated by using gimbals), and therefore “reorientation” can be differentiated as the basic component of precession’s motion (because it can occur without displacement).
The first component (the reorientation of the flywheel) occurs as the torque/force gradually modifies the direction of spin in the flywheel. In my opinion this component of precession DOES requires an energy conversion, in the same manner that energy is required to change the direction of an object moving in a straight line. (A rocket in space uses energy to change its direction.)
The second component of precession involves the displacement motion of the flywheel (along an arc). This displacement aspect of precession DOES NOT require an energy exchange or conversion, even though it is the component of precession that we notice and the one we are most likely to focus on.
I am in agreement that only the “displacement” or translation motion of precession can indeed be the equivalent of a gyro’s stationary position. This is true because the displacement “motion” exists only to maintain the flywheel’s most stable spin (according to the rules of spin), as the applied torque modifies the spin’s orientation.
Note also, the “displacement” or translation motion of precession has a reduced level of equal-and-opposite reaction because:
A) Displacement results only from the reorientation of the flywheel’s mass (therefore…)
B) It requires no centripetal force (the mass of the gyro does not depend on its attachment to the center to prevent it from flying away in centrifugal momentum; if anything, the rigid axel attachment may pushes as much as it pulls)
C) The length of the displacement-radius acts as a lever to magnify any counter-forces, which are created by encounters with obstacles (thus precession seems to disappear when it encounters an obstacle)
The equal and opposite reaction of precession’s displacement only occurs during the brief time that it takes to accelerate from zero to normal precession velocity.
Note also that spin-rate (in proportion to the force of the torque) provides different levels of equal-and-opposite-reaction in precession’s displacement motion. Slower spinning flywheels behave closer to deadweights (and their behaviors can appear more erratic).
In solving mysteries detective often follow money trails. To solve the phenomena of gyros and to find the rationale behind converting angular forces into linear accelerations (gyro-propulsion), we must follow the trail of energy exchanges. The bottom line is that to produce linear acceleration from angular forces, there must be energy conversions or exchanges.
Converting angular forces into “internally generated” linear acceleration will require a number of energy exchanges because propulsion requires energy conversions or exchanges (finding the trail of energy conversions or exchanges is just one step to determine whether gyro-propulsion is possible or not). Whether we are more correct referring to these energy interactions as “using energy” or as “exchanging energy” is a discussion for another day and hopefully another thread.
Most of us know that it is not possible for gravity to deliver sufficient force to lift the mass of the object that is itself creating the force; however, intimate understanding of precession provides invaluable knowledge applicable to power-driven devices, which may in-fact produce sufficient linear propulsion for lift-off (if it is possible).
Best Regards,
Luis
|
| Report Abuse |
| Answer: |
Glenn Hawkins - 05/01/2008 10:11:30
| | |
It was my mistake in creating the term ‘using energy’; as well it was another’s mistake in creating the term ‘absorbing energy’. Both should be abandoned.
The term ‘Using energy’ confuses the simple and is redundant, for its meaning is the same as the standard idea and word ‘reaction’, such as happens in a chain of reactions like billiard balls colliding into one another wherein energy is not used up, but transferred.
Likewise the term ‘Absorbing energy’ is equally problematic and confusing. In the case of ‘absorbing energy’ its meaning applies to equal and opposite forces opposing one another such as happens during precession between angular momentum and gravity and such as happens when you clasp your hands to together and push each against the other with equal force. The old standard and universally understood vectored idea, ‘equal and opposite forces cancel out one another’ should be used instead in both thought and expression. For who am I, and who are you my friend to reinvent the vernacular and meaning of the science of physics? We have been equally guilty.
One other thing is worth saying. If everyone can understand, ‘like a knife through butter, ‘ easy as pie’ Einstein’s words and meanings, Faraday's, Carl Saga’s and Steven Hawking’s, why should I have to struggle with some pieces presented here? The best expository writing is always the simplest. If the writer has literary abilities he can rest assured they will always show as well as a unique perspective of a subject if he has one and without the idea that complexity adds authority. It does not. Really fine work is easy to understand. I am reminded of something expressed by, Sandy Kidd. It meant,” Why belabor the simple, until it becomes complicated.”
I may later address the questions posed here when I have nothing better to do. I know the answers as no doubt so does the asker know all but one. Does it matter? I am praying to do more completely reliable tests. That is so…so hard. If I ever am lucky enough to do and prove another I will tell you about it.
Hawkins on Hawkins
|
| Report Abuse |
| Answer: |
Glenn Hawkins - 05/01/2008 11:28:31
| | | Dear Nitro.
As I said, I like your original post above a lot. I still find it excellent this morning and the ideas …new!
This is what is actually happing during precession. Precession is the reaction to the energy generated, because of the grip and tilting fall into gravity. Precession then causes a reaction itself as it revolves around the string. The reaction is lift support converted into a torque down upon the string. The torque downward is countered by the uplift of the string. The force downward then and opposing force upwards cancel, or nearly cancel out one another. This is how the gyro stays aloft. Now to the full circle of your post, precession does not register energy as you say, because at the same time it accepts energy, it dumps energy into the process of torque. Walla! Reaction-- to reaction-- to reaction. Precession empties itself of kinetic energy as fast as it is filled and so hasn’t any revolving energy left with which to dump into a collision. It seems every bit of incoming energy is redirected upwards before a collision by revolution could take place.
Precession is an empty boll of soup as you all have been telling me all along and as my own tests indicated, but I had to have a mechanical explanation if I had to build the explanation myself before I would accept the idea even though I was fairly sold on it all along. I’m with you now. This has been the mechanical explanation that makes your original post above absolutely fabulous!
Shall I tell you how to capture the rotating momentum--- before it is transferred into torque reaction resulting in upward support? You already know? OK.
Best regards.
Glenn,
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 18/02/2008 19:09:21
| | | Dear Nitro,
I can see how one may perceive precession as the equivalent of standing still. For example precession would go on forever if there were no friction in the spin and in the pivot; where would the energy come from? Also, spin-deflection caused by mechanical torque has a final destination toward a stable position, while precession does not.
First of all, I hope we are in agreement that common earthbound precession can not occur in complete absence of friction, as the gyro pedestal needs to resist the “counter-torque” to gravity’s torque; the counter-torque, which must occur as a reaction (not in response) to the force of the torque that is necessary for precession to occur.
Using a spring attached to a significant counter-mass, one could replace the torque of gravity in space and thereby replace the need for friction with a counter-pull on the counter-mass as the spring tries to bring the counter-mass and the gyro’s flywheel closer.
In other words, away from earth’s gravity (in space) friction may be made unnecessary by using a spring (in place of gravity), and if a significant counter-mass is used to supply a degree of anchor to the other end of the spring, the spring can produce a proportionate pull on the spinning flywheel thereby producing precession (this device is relatively easy to build and test even on earth).
Precession, in such an apparatus, will come to an end when the force of the spring torque brings the counter-mass very close to the mass of the gyro flywheel (i.e. the spring is fully contracted). A normal earthbound gyro is not able to move the entire planet (i.e. the counter-mass) toward itself, but that does not mean that it is not trying to do so.
I also think that the fact that gyros move around without a final destination does not qualify their motion as the equivalent of “standing still”. Perhaps I should ask whether mechanical driven spin-deflection should also be considered as “standing still” even though it has a final destination?
It’s fair to add that if a force does not move an object, it does not perform work, and things are standing still. Thus we may say that when the torque that causes precession fails to move the planet (its counter-mass), it does not perform work either (bear in mind that precession is a motion that arises from the changing direction of another preexisting motion, the spin).
Interactions of massive objects, with objects of relatively small mass eliminate the effect of the small upon the large. In such cases 100% of the accounting is taken as the effect of the large mass upon the small one, and zero effect upon the large mass (and this produces results that are close enough to accommodate common experimental results).
It’s difficult to determine the exact “cut-off” proportion, where one object’s mass becomes 100% dominant; however, there is an equation in this forum to calculate the impact between masses, which may serve as a good proxy.
If one subscribes to the belief that the smallest object maintains an inordinately small, but still existent effect upon the largest object, then precession’s effect on the planet becomes a matter of degree, however small. If we adhere to this belief then this smallest of effects is still a significant component that contributes to precession.
There is no doubt that precession requires a freedom of motion that does not actually take place, and that missing motion would be the attraction between the flywheel and the planet.
Precession’s force is different from centripetal force eave though both may simply exist in a “potential” state but without actually creating motion an externally relevant motion.
We may chose to compare precession to a simple spinning object because it too has circular motion and involves a force that does not produce work. In this regard, I can agree that precession is a form of standing still, in as much as an object spinning in space can be said to be standing still, but not beyond that.
Best Regards,
Luis Gonzalez
|
| Report Abuse |
| Add an Answer >> |
|
