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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.
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Momentus |
| Subject: |
A working model |
| Question: |
I have been reading thro’ the various posts on the site and decided to share my experience with gyroscopes over the past 30 years.
My basic model is a gyroscope suspended by a long light cord.
It consists of a flywheel at one end of a shaft, with a compensating weight at the other end of the shaft and a cord attached to the middle of the shaft.
It has performed consistently over the years, with a variety of wheels and forms the focus of my understanding. The behaviour of this model challenges classical physics.
Critical to the performance of a gyroscope is its efficiency. A gyroscope of 100% efficiency has all the mass distributed at the rim of the wheel; this gives a maximum value for the moment of inertia (M of I)
By contrast a disk has an M of I equal to ½ of this value. To obtain the same gyro effects of couple and precession would require twice the mass, giving an efficiency of 50%.
This extra mass is present to some degree in all real world gyroscopes and is detrimental to the overall performance. In the basic model it is compensated by an additional mass placed at the opposite end of the gyroscope shaft. The mass required can be calculated from the ratio of the actual M of I to the ‘perfect’ M of I, or by the method shown, adjusted empirically.
When the spinning gyroscope is released, the basic model precesses about a centre of rotation located at the point of suspension, midway along the shaft. If the centre of rotation does not exactly coincide this point, it can be adjusted by altering the compensating weight.
When the centre of rotation coincides with the point of suspension and the cord is vertical, by the simplest of Newtonian mechanics it can be proved that there is no force in the horizontal plane.
The mass is orbiting without a force to cause the inward deviation from a straight line. No force means that there can be no centripetal acceleration.
As the mass of the model is offset from the centre of rotation, that produces an action which transcends current understanding.
Taking a viewpoint from above, the gyroscope will precess in a circle. Assume that it is released at the twelve o clock position and allowed to rotate around to the six o clock position; it will be displaced by the diameter of a circle that is centred on the point of suspension.
The point of suspension is also the internal/external interface. As already stated, the vertical cord plus simple mechanics proves that the mass of the gyroscope is moved without external action in the plane of motion.
Displacement without a force along the line of action of the displacement also transcends current understanding.
By simple mechanical ingenuity, using an efficient gyroscope this displacement thro’ an arc without force can be made continuous.
Placing the gyroscope in the centre of the shaft, and alternating the suspension point from end to end of the shaft, will cause the gyroscope to ‘walk’.
These are not proposed experiments they are current working models.
This transfer of mass can be seen in the Alex Jones video, and in the Prof Laithwaite patent.
The walking gyroscope will only move its own spinning mass. There is no force present to overcome any impediment. Trying to pull a light trolley for example results in the walking gyroscope oscillating without forward motion.
Once the walker has moved itself over a given distance, it can be returned to its point of origin by simple mass transfer Newton style. If the Walker has moved itself from A to B, an equal and opposite mass can be moved to B by returning the gyroscope to A.
This behaviour does NOT fracture the Conservation of Momentum law.
That will do for now, I have no idea whether I have made this clear, so I will await feedback before trying to take it any further.
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| Date: |
11 February 2005
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| Answer: |
Sandy Kidd - 14/02/2005 07:59:05
| | | Momentus
Looks like there are at least two of us flogging the same line.
Apart from the last part of your posting relating to Laithwaite / Jones type mass transfer of which I have absolutely no experience, I would suggest your findings and claims are pretty much in agreement with my own, and for the reasons below.
I return to my original claim that any mass, subjected to radial acceleration, which is itself rotated (i.e. flywheel or gyroscope), the system will start to shed angular momentum / centrifugal force ( no apologies for this) as soon as rotation of the mass (flywheel / gyroscope) commences.
This shedding of angular momentum / centrifugal force (again no apologies) will continue right up to the point, where so called “precession” occurs.
It is therefore fairly obvious that gyroscopic mass will also be proportionately transferred.
This so-called “precession” only occurs because there is nothing left to shed.
At this point there is nothing left in the system but rotation itself, and an inward acceleration of the flywheel / gyroscope, due to its torque reaction.
Takes a little bit of getting the head round this phenomenon, but that’s the way it is.
Experiment
Take a pair of gyroscopes in an opposed balanced system.
Run the system at a fixed speed without gyroscope rotation, and it will be balanced.
Run the system at the same fixed speed, with both gyroscopes rotating at equal, but any speed and the system will remain balanced.
Now gradually decrease or increase the speed of one of the gyroscopes, and immediately a mass imbalance will appear in the system, very much akin to the effect, a non rotating mass, offset by a gyroscope, will display.
Always wanted to do this with strain gauges suitably attached.
The point is that nobody wants to believe it anyway.
Warning.
If your system is being rotated at a reasonably rapid rate of rpm, do not overdo the gyroscope speed differential, the outcome could be very dangerous.
Sandy Kidd
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Victor Geere - 15/02/2005 11:23:28
| | | The explanation is very clear thank you.
I am not convinced that there is no horizontal force on the balanced gyroscope though.
If the shaft instantly turned to rope and the gyroscope stopped, the gyroscope (and the balancing weight) would drop at an arc and hang straight down below the rope that suspended the shaft. The effect of gravity is therefore an arc between the vertically hanging gyroscope and the horisontal spinning gyrosope.
Back to the original shaft:
When this gravitational arc is changed at a 90 degree angle by the effect of the spinning gyroscope, it equals the path that the presessing gyroscope follows around the suspending rope.
This horisontal force arc is the gravitational arc at a 90 degree angle.
I agree with the observations that you made, this is just an attempt at guessing where the force came from that moved the gyroscope from 12 to 6. I look forward to the rest of your article.
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Momentus - 15/02/2005 14:11:18
| | | Hi Sandy
Have been giving considerable thought to your post. We certainly have agreement on the basics. Gyroscopic anomalies do exist and can be exploited.
The absence of centrifugal, centripetal or any other acceleration is a further point of clear agreement.
Your use of the word precession troubles me. >>….. will continue right up to the point, where so called “precession” occurs. <<
I take it as given that in all cases the gyroscope is spinning. From this, any angular motion is precession, and has an accompanying gyroscope couple.
Taking the case of your model with opposed offset wheels (horizontal), which you start to rotate about a vertical axis. There is no opposing force to this precession; from the first detectable movement it is the precession/couple that drives the model around the vertical axis. The angular momentum of the flywheels is reduced by the vector sum required to produce the motion about the vertical axis.
The action of increasing the speed of vertical rotation is also the same action as increasing the couple and occurs simultaneously.
So I cannot grasp your concept of a point when precession ’kicks in’. Precession/couple is one action and is always present.
You talk of shedding angular momentum. Is this the same thing that I consider as the sharing of momentum between the horizontal and vertical axes?
To precess a gyroscope takes no input of energy. No work is done. No external force is present.
Within these constraints, I can comment on your first lifting model, relating the actions to my more basic walker.
As the flywheels lift, whilst precessing, they trace a shallow upward spiral. This requires no additional vertical acceleration. It is a zero force motion.
From the top of its lift, the wheel is jerked down linearly giving an upward pulse.
This gyroscope in one direction, inert mass in the other, is the same action as the walker uses. Like the walker, no momentum is created.
I think that this action came from the bearing slogger you refer to. The distances moved would be small, but rapid.
Finally, the demonstration precessing with one gyroscope spinning, and one stationary, spectacular. How can that possibly be ignored??
Do have difficulty with >> decrease or increase the speed of one of the gyroscopes. <<
The gyroscope couples would be unbalanced, but not the centrifugal forces, which are non-existent regardless of speed. So is it the couple that tears it apart?
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Sandy Kidd - 16/02/2005 08:09:53
| | | Good day Momentus,
Thank you very much for your comments, and bringing a point, that is a continual irritation to me into the conversation..
In the context in which I was describing the operation of accelerated systems, you made mention of precession (and forced precession)
I wish this terminology would just disappear altogether.
Precession should be applied to passive systems only.
There is no precession in an accelerated system.
We are unfortunately forced to used these words to describe what folks believe they mean.
The erroneous point of precession in an accelerated system does not exist as such, but when the gyro overcomes its accelerated mass due to its torque it is commonly and totally wrongly called precession. That is why I call this point the saturation point (or gyroscopic torque saturation point). No increase in system rotation speed or gyro rotation speed will affect the system once the system has reached this point. The only change will be in the increased inward acceleration rate of the gyroscope.
So, I apologise for using the word precession, which it is not, but how else do you describe this effect which most people have never seen, and most find it hard to visualise
In future I shall refrain from using this kind of terminology when describing accelerated systems, and I shall also use the term “flywheel” instead of gyroscope.
Howsoever I must beg to disagree with your comments relating to centrifugal force.
If the system could instantaneously reach saturation rotation speed at the instant the flywheels reached saturation speed, I might agree with you.
The only point at which “all” the centrifugal force in an accelerated system is neutralised is when the system has reached the “saturation point” and at this point I hasten to add it is no longer of any use to us inertial drive freaks.
The only reason the flywheel remains where it is, because even close to the saturation point it still retains some mass, which can be accelerated. Only when all the mass has been effectively transferred, and there is none left to accelerate, does the flywheel begin its inward acceleration, the path of least action.
Remember I am not describing a passive system.
In the University of Dundee Laboratory, many moons ago now, I actually created a system, where a flywheel could clearly be seen to oscillate, at a rather leisurely rate in its orbit so to speak, by means of a rather expensive strobe light, University equipment of course, and a rather cunning, (well I thought so), flywheel rotation speed changing mechanism..
I was placing the flywheel into saturation and bringing it out again.
This for your interest was in a balanced twin flywheel system (what else?)
Funnily the boffins did not think this was worthy of note. This was fair enough I suppose as they were much more interested attempting to prove the device was a fraud.
They never did of course, but it did not matter, it could not possibly be genuine, could it?
So it was out.
That’s science, and c’est la vie.
Sandy Kidd
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Sandy Kidd - 16/02/2005 12:38:19
| | | Hello again Momentus,
Could have carried on in my previous posting but I would have got carried away completely.
Anyway I have returned to put the record straight.
You made mention of my original machine, and you made a reasonable guess as to why it worked. Unfortunately it was not that simple.
I had thought out a way of creating inertial thrust many years previous to attempting to try it.
I called this my “Bricks on Strings” theory which I was tempted to post in the Forum some time ago. Howsoever common sense prevailed and I desisted.
This was based on physics and mechanics, the way I had been taught at school and at technical college.
Only after I built and tested the machine did I discover that my understanding of gyroscope physics was wrong. I pursued the problem only to find out that what I believed was common belief and I was not misunderstanding what I had been taught.
This made a complete mockery of my grand idea.
To cut a very lengthy story short, the machine worked very consistently, but for a whole load of reasons I was not aware of at the time.
I then for some reason believed there had to be an oscillating component in there somewhere that I could not see, as the machine operated as smooth as silk.
Thought maybe Scott Strachan was on the right track with his cam operated device. I made a cam-operated device that worked quite well, but this was obviously not the way the first machine was creating its thrust.
Went through the “free lunch” thing you suggested and the answer was definitely not in there.
It took me the best part of 12 years to figure it all out, the trouble is that it could be a mechanical nightmare to reproduce the effect. It is one of these peculiar effects that happens, as a result of cascading conditions. A variety of different elements were present, which when combined in a specific pattern made the thing work.
The mechanics of the machine required to produce thrust is simple enough, in fact I built up a new and complete machine suitable for the purpose over the weekend. Luckily I managed to steal parts from other machines to do the job.
The rest is not so easy. I have a few options on how to attempt this, one is simple and has been previously proved but is very expensive. Any kind-hearted person got a spare Mori Seiki NC lathe I can borrow? None of the other options are simple, but are much less expensive. In the meantime I will be carrying out a few relatively simple tests.
Sandy Kidd.
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John Entwistle - 22/02/2005 07:45:38
| | | Here's THE answer bro there's nothing new under the sun perhaps ovah the moon
this is my idea soley
29"bicycle rim inflatable exercise ball put these 2 simple objects together fill BALL with
^(^( NANO gyroscopes attach to any steel 2 gryoscope driven object have fun go skiing
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Momentus - 23/02/2005 20:20:49
| | | Hello again Sandy
I’ve been doing a lot of thinking since your last post, whilst popping back to the BBC boards, just to keep things stirred up and at the same time moving all my programs and data onto my new computer.
What really got the little grey cells agitated was your “forced to use these words”, something that I have also raged about. The vocabulary of mechanics and physics has evolved to fit the classic paradigm. New concepts need a new vocabulary.
I then realised that I was looking at your posting in exactly that way!!!!!! Of course I do not understand what you are saying if I translate it back into “classical” in my head.
The systems we are talking about have a spinning flywheel on a shaft, which is rotated, usually about the vertical axis. You use paired flywheels, force the axial rotation and react the gyroscope couple by keeping the flywheel vertical.
Gyroscope precession is the rotation of a spinning flywheel about its C of G. It is accompanied by a gyroscope couple. These two elements are orthogonal with the spin axis.
An offset gyroscope has third element, linear motion. This linear motion is tangential to the circle described by the offset gyroscope.
By combining this linear motion with precession rotation, the gyroscope follows a circular path without centripetal acceleration.
Imagine an ant crawling along the rim of a hex nut. It turns at a corner, and then goes along the straight, turns etc. If it takes one minute to turn through 360 degrees, the linear speed along the straights of the hex determines the size of the hexagon. Increase the number of sides until it becomes a circle. This is an approximate explanation.
For a passive gyroscope this circle is coincident with the pivot point of the torque arm. Increasing the linear speed by increasing diameter of the circle also increases the torque, which increases rotation and tends to be self correcting. Thus the gyroscope on a string has no force acting.
Which I assumed would be the same for all systems, including forced axial rotation. Sandy says that this does not apply to his forced axial rotation. Even more than that, he gets an inward acting force, which varies with rotation speed.
I need to explain, which is where vocabulary and concepts come in. Try to look thro the words to the meaning.
The gyroscope on a string follows a neutral circle. If the combination of torque arm length, spin speed and precession required it to move in a larger circle, then a force would need to be present to move it into that larger circle. Once it had reached this new circle, it would be balanced again. There would be no force. The new circle diameter would be the neutral circle for those conditions.
So by changing the rotation speed in a forced axial system, but not the length of the torque, the conditions for a new neutral circle can be set up, but the flywheel can be prevented from moving to the new diameter. A force exists.
Which at least answers one of my own questions, as to how rotating paired flywheels at different spin speeds can give unbalanced forces?
The acid test of any theory is a prediction.
Sandy, you say the flywheels move inwards. Do they move to a stable position, or once started, do they keep on going?
Ill sign off with that
Momentus
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Sandy Kidd - 24/02/2005 12:37:20
| | | Hello Momentus.
Liked your ants and hex nuts.
In a system subjected to radial acceleration, flywheel torque is a product not only of flywheel rotation speed but also of system rotation speed.
The flywheel is acting somewhat like a “spanner in space” the turning force of the spanner being proportional to the flywheel rotation speed, that is of course assuming the system rotation speed is constant.
In a forced system, when the flywheel rotation speed has been raised to a point, sufficient to overcome its own accelerated mass, it will continue to accelerate in an arc, upwards and inwards, around its fulcrum. This is in an attempt by the flywheel to gain a position where its axis of rotation is in line with the axis of system rotation. This is the point of least action.
In a forced system with a fixed rotation speed, the flywheel will obviously develop no torque at zero flywheel rotation speed, but will develop maximum accelerated mass.
As the flywheel’s rotation speed is increased, torque will correspondingly be increased.
The flywheel torque creates an inward acceleration of the flywheel, offsetting the “wait for it” centrifugal force. So it can be seen that the centrifugal force in the system can vary from all, to nothing at all, depending purely on the flywheel’s rotation speed.
Sandy Kidd
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Momentus - 08/03/2005 11:44:01
| | | Hi Sandy
I am Still struggling with vocabulary. For a wheel to become a gyroscope it needs to satisfy the conditions for the three orthogonal axes. There must be a spin axis, a motion axis and a force axis.
The Force produced by a gyroscope, whether it is in gimbals or is offset, is a couple it is not a torque.
A couple is equal and opposite parallel forces. This will result in rotation about the mass centre. A torque is one force acting at a distance from the mass centre.
The gyroscope couple is reacted by a torque, when the shaft bears down on the outboard pivot. If the couple is greater than the torque, then the flywheel will move.
The couple/motion action is an instantaneous thing. Precession forced or otherwise neither precedes nor lags the couple.
The saturation point you refer to seems to me to be when the gyroscope couple exceeds the reactive torque from the weight of the gyroscope. Lay the apparatus it on its side and the wheels will move into alignment with a minimum of forced precession, unless you have some springs or pistons to exert the reactive torque.
Torque is a spanner, couple is a Tommy bar, and either of these will lift a dead mass at the end of a shaft, no gyroscope needed.
I am missing something here, but what?
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Sandy Kidd - 09/03/2005 07:19:00
| | | Hello again Momentus,
Will be easier for me to start at the end of your statement,
Yes a spanner or a tommy bar will exert torque but in both cases require external leverage against the nut it is tightening.
I wrote “spanner in space” to describe gyroscopic torque effect, nothing else.
Torque means turning force however inspired.
A gyroscope does not require this external reaction, it does its thing quite happily by itself as long as it is accelerated.
This is what makes them the main contenders for inertial drive.
Now to this couple you mention in relation to the gyroscopic torque?
Am I to believe that the top sector of the gyroscope is trying to turn inwards around the centre of mass of the gyroscope, as you imply, and the bottom sector turning outwards with equal thrust to create this couple?
Could you please enlighten this old man as to how the top sector of the gyroscope achieves this inward thrust?
Could you also enlighten me as to who ever implied that this was a couple in the first place?
At the saturation point I keep on about, the gyro is overcoming copious amounts of accelerated mass and a small amount of rest mass.
This is not a passive system. The rules are completely different.
Gravity is not needed in an accelerated system, in fact it is not even wanted.
In all of my machines the accelerated mass is many times that of the rest mass.
In other words the affect of gravity on the issue is minimal at these elevated speeds. The faster the machine runs the less it matters.
Sandy Kidd.
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momentus - 17/03/2005 14:54:56
| | | >> Torque means turning force however inspired<<
There is a big difference between couple and torque. I use the Tommy bar/spanner to illustrate this. The Gyroscope produces a couple internally, not a torque. To convert this couple to torque an external reaction point is required.
>> Could you please enlighten this old man <<
We examine this problem like a bunch of blindfolded experts examining an elephant. I talk about the tail, you are describing the trunk. Same beast.
Many years ago, I read the book ‘Gyrodynamics and its engineering applications’. It advises the reader not to attempt to understand how a gyroscope works, but to rely upon the mathematics, and then came some 500 pages of math and formulae.
In the late 1980s I had the privilege of meeting with one of the authors. The purpose of the meeting was to obtain the professors opinion on the power transmission system that we had patented. Using only the maths in his book, and a model of the apparatus to show what it could do, I received a glowing endorsement. Not given casually I might add. He was extremely well versed in his own formulae. Mind like a steel trap. Great meeting.
Two weeks later the endorsement was withdrawn. No reason given, no further contact permitted.
So there was more to it than the math. I needed a visualisation, some way to see what was happening. This led to my perfect circular billiard table.
This is how a gyroscope works.
Imagine a circular billiard table, with one ball (any colour ) bouncing from cushion to cushion over the centre spot. As it is a perfect system, this motion continues for ever.
Now as the ball passes over the centre spot, deflect it by a small angle, as it returns, give it an equal and opposite deflection. Repeat ad nausea. The ball will circle around the rim of the table.
Once you are happy with your visualisation, change the table for a crystal sphere. As the ball passes the North Pole, deflect it, as it passes the South Pole, equal and opposite deflection. The ball will circle around the equator of the sphere.
This illustrates the three orthogonal aspects of a gyroscope.
1) Spin, the ball bouncing from cushion to cushion.
2) Precession, the ball circling around the rim.
3) Couple, equal and opposite deflections
>> Could you also enlighten me as to who ever implied that this was a couple in the first place? <<
It is a couple, equal and opposite forces. I figured it out for myself.
>> At the saturation point I keep on about, the gyro is overcoming copious amounts of accelerated mass and a small amount of rest mass. <<
I still have not grasped the saturation concept, but I live in hope.
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DaveS - 17/03/2005 16:50:49
| | | I never bothered to immerse myself in these mathematics. I did study some of it initially but found that there were too many assumptions that do not actually explain a great deal when looked at closer. In addition, I did not find it relevent in relation to my design. If you have also got your head around it then i applaud you, it's heavy s**t that most are unlikely to attempt to decipher. I strongly suspect that your mathematical theorum genius removed his endorsement because he knew you would end up picking holes in his logic or that his logic would not apply to gyroscopic propulsion.
I can understand the logic in trying to understand what is going on but to apply numerical solutions without a complete formula is a nonsense.
As i keep saying, "Do you know how a wheel works?" If you do, how do you know? Have you ever sat down and found it necessary to calculate pi and apply numerical values to justify how the wheel works? If you have not and you know how the wheel works, then why try to apply mathematics to gyroscopic propulsion? It is not exactly the same as nuclear physics where theorum is extremely important prior to experimentation. Especially when the experiments and the measurement techniques are often dubious to say the least.
Interesting article though Momentus.
DaveS
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Sandy Kidd - 18/03/2005 08:14:20
| | | Absolutely spot on Dave S
There are just too many assumptions relating to the physics of gyroscopes, most of them wrong.
It is very obvious that very little experiment has ever been carried out.
Simple passive systems, for instance have been researched to the nth degree, and they still do not know what is going on.
By the way Momentus, saturation to an accelerated system, is what precession is to a passive system.
And yes Momentus I originally accepted the fact (many years ago now) that it was a couple because someone said so, and it did not affect me too much.
Think I read it in a book.
That is a typical example Dave, assumption.
Would someone please explain to me, how it can possibly be a couple.
Does not affect the outcome anyway, but why guess?
Momentus has somehow convinced himself, wish he would try to convince me.
Then we have the coup de grace.
As soon as you start to accelerate them (gyroscopes), the whole issue goes totally pear shaped.
We have disappearing angular momentum and centrifugal force.
This is why I still use the term centrifugal force.
Centrifugal force, changes to centripetal force at the point of saturation.
We can also have transfer of mass, at any rotation speed.
A rotating ball will rotate inwards during acceleration.
Centrifugal force can be switched on and off almost instantly at will.
And there is a lot more besides.
Find any of this stuff in a text-book if you can?
Fortunately from a personal point of view I am not too much interested in anything else except maybe, aerodynamics, and I shall say no more about that.
I sincerely hope the rest of our physics is in better nick, than the trash that relates to gyroscopes.
Really wound up now
Sandy Kidd.
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arthur dent - 06/04/2005 17:52:34
| | | An interest in aerodynamics, eh. I think that we may have a point of agreement here. I bet that you believe the Bernoulli explanation for flight (air travels faster over the top surface, etc.). And here is the point of agreement: this is certainly a case where the textbooks are indeed full of drivel. If the Bernoulli explanation were correct, an aerofoil could be turned into a reactionless space-drive. But don't bother to patent the concept; it has already been done.
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Sandy Kidd - 07/04/2005 15:25:16
| | | No iI don't.
Where are you going to find air for your spacedrive.
Real clever stuff.
Way beyond me, I'm afraid
Sandy Kidd
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arthur dent - 08/04/2005 00:48:09
| | | I did not say that it would work; only that the popular (and wrong) Bernoulli explanation for flight does not rule it out (it is thus a sort of physical reductio ad absurdam). The concept has, however, been patented. The air is contained within the body of the space-drive, and is endlessly circulated. Now, is that any more ridiculous than other ideas mentioned in this forum; concerning jets of fluid, 'one-way' impulses ... and gyroscopes that react against nothing?
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Sandy Kidd - 08/04/2005 08:22:03
| | | Arthur Dent,
If that is your real name, as you now inform me, I apologise for assuming you were the last man alive.
A most unfortunate coincidence
By the way I am an engineer, not an inventor.
However you skipped conveniently past my statements relating to changes of angular momentum, and other factors relating to accelerated gyroscopic systems, without making comment.
You picked up on my interest in aerodynamics, and made comment referring to the out of date aerodynamic theory relating to lifting wing sections, a fact that been common knowledge for many years (but I hasten to add is still taught)
Similar in many ways to the gyroscope we have managed to live without the truth for many years, and again like the gyroscope, your red bricked bastions of credibility, were/are handing out diplomas in obsolete theory, long after the truth was/is known. I do not know how the system works but are all these “funny” diplomas withdrawn/cancelled/annulled/whatever?
You did skip my statements relating to angular momentum changes etc. without comment.
Can I take it from this, that you have no disagreement with any of my claims?
Sandy Kidd
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arthur dent - 13/04/2005 21:35:31
| | | Just to be clear: the old theory of flight, the one which was assumed by all of the aviation pioneers, is entirely in accord with Newton's third law. The Bernoulli theory, which has become so popular, is not. I would even agree that modern textbook accounts of gyroscope theory are incomplete. Not wrong, but incomplete. That is not to say that there is anything new to learn but, rather, that a lot of things which were known about gyroscopes in the 19th century have been forgotten. For instance, it is a very little known fact that it is possible to rotate a rotor through a finite angle, without the rotor ever acquiring any angular momentum about its axle. (I could explain here how to do it, but I am thinking of writing an article about it for a physics magazine and I do not want to spoil the surprise). I have not commented on your angular momentum ideas because I agree with some of them (when they coincide with fact) and not with others. As this dichotomy sometimes occurred in the same sentence, it was too tedious to comment at all. But, to pick out 2 points: it is nonsense to claim that a precessing top exerts no 'centrifugal force' (to use reluctantly that dubious concept) about its support, or that a precessing top has no angular momentum about the support.
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Glenn Hawkins - 06/07/2006 13:13:31
| | | Dear Arthur,
You are talking about rotation. With a spinning top one can conceive centipede, with precession one cannot. I in unpopularity here agree with you that there is centrifuge and momentum, except that as I understand it centrifuge is compensated for in a complicated way so that it dose not exhibit itself in a visual, or measurable way. A top is not a good analogy, because it's spin is balanced by equal and opposite masses traveling in opposite directions. Rotation can be two balls tied to a string rotating in a binary orbit. In precession the condition is singular. There is no opposite mass traveling in an opposite and balanced direction held together by means of centipede. A spinning top dosen't precess.
Glenn,
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Glenn Hawkins - 06/07/2006 13:55:24
| | | My mistake, Arthur. Excuse me. I see how you’re relating a top leaning at an angle above the floor and the action being the same as if it were a thin flywheel instead of a fat one. Duh!
Momentous, I found your better explanation here. Good show. It’s theory, and the actual actions will be short lived. It couldn’t work continuously without special alterations. Some things are missing, but again, I think it is an excellent beginning and good work.
Glenn,
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