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29 November 2024 00:46
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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.
You may also be able to help others by answering some of the questions on the site.
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Question |
| Asked by: |
Luis Gonzalez |
| Subject: |
Third Law of Motion |
| Question: |
The 3rd Law states that for every Action there is an Equal & Opposite Reaction.
"EQUAL", means that the magnitudes of the Action and of the Reaction forces are the same.
"OPPOSITE" means that the touch-point of the Action & Reaction occur at 180 degrees from each other.
Nothing in this Law states that both the Action and the Reaction need to be linear (in fact the 3rd Law has been extended to include angular Action & Reaction).
Can a device produce Linear Action with Opposite Reaction driving angular motion (i.e. rotation)?
EXAMPLE:
1 - A spoke on a flywheel uses a strong spring (or a small explosion) to fire off a mass in a direction TANGENT to the flywheel's perimeter.
2 - The ejected mass travels in a straight Linear trajectory.
3 - The Reaction force accelerates the mass of the device into an Opposite Linear direction but most important it also accelerates the mass of the flywheel into Angular Rotation.
Does this break the 3rd law?
(Net magnitudes are same, and are opposite at the touch-point.)
Best Regards,
Luis G |
| Date: |
15 April 2012
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Answers (Ordered by Date)
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| Answer: |
Luis Gonzalez - 15/04/2012 15:52:29
| | | Also, does the sum of the linear and angular reaction forces in the example equal the force of liner action of the ejected mass?
Luis G
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| Answer: |
Hatry K. - 18/04/2012 20:29:05
| | | Hi Luis,
The sum of linear and angular reaction energy should be equal to the energy stored in the linear ejected mass. The most part of reaction energy will be transformed in angular momentum.
However, t
his is my understanding of physics.
Best regards,
Harry
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| Answer: |
Glenn Hawkins - 18/04/2012 21:22:19
| | | Hello men,
A paddle wheel is in space. A meteorite strikes the face of one of the paddles then bounces backward. The result is that the paddle wheel. . . . . .
SOLUTION
Lay an ink pin on the table and flip one end quickly with your finger and you will see your answer in action form. Forget about friction. It will have only a negligible effect.
Glenn,
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| Answer: |
Luis Gonzalez - 19/04/2012 03:24:41
| | | Harry, Thank you for confirming that a portion of a linear motion's magnitude can be an angular opposite reaction.
So, it looks like linear actions with angular reactions do not break the 3rd law.
However, since the angular reaction is internal to the device, an external observer would think that the linear action (the ejected mass) occurs without an EQUAL and opposite reaction.
The external observer would be in error because he fails to see the complete sum of the reactions in the example presented.
All other thoughts are welcome and highly encouraged.
Best Regards,
Luis G
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| Answer: |
Glenn Hawkins - 19/04/2012 04:39:09
| | | Beautiful Luis, but the the external observer would be correct relative to his position in space. That is what relativity is all about. Aw, but you knew that.
Glenn,
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| Answer: |
Luis Gonzalez - 21/04/2012 03:05:34
| | | Hi All,
I would have preferred an expert on Relativity to set the record straight (Ravi is probably otherwise occupied). Glenn, Thank you for the compliment and I mean no insult by what I am about to say.
Let's clarify that to laymen the term "relative" is applicable to differing perspectives about any set of relationships or interactions.
In contrast, Einstein's concept of "Relativity" refers strictly to the time-space continuum, as units of measure become affected proportional to the speed of light. (Does Ravi disagree?)
The external observer (in this thread's example) does not know (or does not understand) the inner workings of the device. This one specific missing piece of information causes his incorrect conclusion.
It has nothing to do with space-time continuum nor distorted units of measure, in either frame of reference.
In fact, the external observer can have the device resting on the palm of his hand, and still make the same error by not knowing about the internal workings (the observer would still swear that the device produces action without reaction, and that it breaks the 3rd law).
Sorry for the lengthy rant on relativity.
I am much more interested in showing that the 3rd law remains applicable even after an inertial propulsion device is successfully built.
My "rant" is only intended to prevent an erroneous belief that I intended my explanations about the 3rd law to be somehow connected to relativity, as that would make my explanations unacceptable to relativity experts.
Relativity is one of the most misunderstood subjects, next to quantum theory.
I guess gyro dynamics comes a close third.
My Best Regards,
Luis G
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| Answer: |
Harry K. - 21/04/2012 12:24:23
| | | Hello Luis,
Unfortunately I'm not an expert in Relativity but let me add my two cents regarding this issue.
The Newtons 3rd law refers (ONLY) to forces in an inertial system, i.e. a non-accelerated system at rest. However, a rotating system cannot be an inertial system because inertial forces and thus fictitious forces occur in such a system. Thus the 3rd law does not apply in your given example, independent from the location of an observer.
Strictly speaking, there is no inertial system, because, so far we know, everything is in rotation:
the Earth, the solar system, the Galaxy, etc., all is in rotation and thus there is no inertial system at rest at all, at least nobody knows.
What are you diving at? Do you have a concrete idea or is this only a philosophical question?
Have a nice week-end!
Harry
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| Answer: |
Glenn Hawkins - 21/04/2012 13:14:51
| | | It is very simple. What is true in one frame of reference may not be true in another frame of reference. What is true to the observer is true to him. Cause and effect—the reason why, is yet a completely different issue. He sees what he sees as an observer, period. In this case he must leave his reference, which is true to his observation and enter an interior reference to discover the mechanical of cause and effect. The mechanics of relativity are not complicated and there has never been an ‘expert’ tag of reference given to it. It has been explained with such great simplicity that even a dullard can understand and it requires no champions to rant the virtues of its theory against other people. Take a Valium. Only the math is complicated.
Harry hi, I thought it was as a philosophical mind twister, hence my response. I didn’t know he was serious. Thank you for this, “. . . there is no inertial system, because, so far we know, everything is in rotation:”
Glenn,
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| Answer: |
Luis Gonzalez - 21/04/2012 16:52:25
| | | Glenn, It is so nice that everyone in this forum (perhaps in the world) understands relativity thoroughly and finds it simple.
I suppose only dumb scientists believe there is any complexity to it, and the paradox of the twins is really fully understood by all as well.
Harry, Thanks for explaining that the 3rd law never really applied to reality. I guess it probably should have never been taught in school. All the physics course problems were of no help to anyone.
The problem/example I presented in this thread is completely useless. I guess I should have never introduced it for discussion, as the 3rd law is only a figment of the imagination... oh no...not any one else in this forum, only in my imagination.
Concrete idea?? How about Gyro Propulsion, as in this forum? Is that concrete?
Does anyone else care to weigh an opinion?
Regards,
Luis G
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| Answer: |
Glenn Hawkins - 21/04/2012 18:00:17
| | | Hi Luis G.,
Wow! So the world of physics needs a graphic designer as an angry attack dog against kind, genial and agreeable people? Now Listen. Anyone with normal intelligence, or nearly, will be able to understand the basics of relativity if they will try. I'm going to put my foot in my mouth, (I should know better), but go a head Luis and tell me what it is you don't understand. Einstein’s postulations and stick drawings are the simplest in the world. He was a genius for reducing to the simplest. A two thousand year old axiom call Occam's razor, simply states that if there are more than one unusable equation, the simplest is correct one. Einstein said he redid his general theory relativity, because of that axiom requiring things be made as simple as possible. Simplify, simplify, simplify. That you understand the basics of relativity and four dollars will buy you a cup of Starbucks. Maybe the coffee sever can explain it to you. Take notes! If someone wants to study and learn it they will. It has been presented by calm genius' in a thousand books all over the world. I hope you don't shoot anybody today. I love you, Luis.
Glenn,
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| Answer: |
Harry K. - 21/04/2012 20:10:01
| | | Hello Glenn,
Please cool down! Whenever someone wanted to discuss some interesting issues, the thread will be polluted with unobjective postings. I made these experiences in my threads as well. We should respect opinions and thoughts from others to come along with our own understanding of our perspectives.
I know you are an impulsive person and you do not want to insult anybody but you should take care of your word choice. Written words can be misunderstood.
Have a nice Sunday!
Harry
Hello Luis,
It was not my purpose to insult you in any way. I read your interesting question in your thread and thus I gave my two cents to it.
The question was, if the Newtons 3rd law may be broken by your presented example, and I answered with a "no" because your example does not occur in one inertial system, but Newtons 3rd law refers only to an inertial system.
But I also concerned that there is in reality no inertial system present because -so far we know till now- everything is in rotation and thus no inertial system can be present, and in turn the 3rd law cannot be correct. In this context you are right that the 3rd may be broken, but not only in your presented example but rather it may be broken at all.
At last this could be the reason why we waste our precious time here in this forum, couldn't it?
Anyway, I believe to know why you keep yourself busy with this interesting question.
To achieve inertial propulsion, you have to transfer full or at least partial angular momentum into linear momentum. You can do this either if you could achieve to eliminate one force vector of a torque, e.g. how Sandy has done it with his prototype within his saturation zone or in the way I had described it in my balancy point thread, or if you achieve to transfer linear reaction force (i.e. energy) into angular momentum (i.e. rotation energy), e.g. as Nitro has done it with his setup of mass displacement.
Your example discribes a possibilty to partially transfer the reaction force into angular momentum by combining an (quasi) inertial system with an non-inertial system (rotation) to achieve at least inertial propulsion in linear direction of the action force.
Yes, I believe this could be possible due to the gyro behavior to transfer linear acting forces into angular momentum.
If you find a good technical solution you are the winner! Nitros and Sandys approach are promising. . Hawever, I have my own ideas in this matter as well and I'm sure you too...
Sorry again to break in your thread. Please let me know if you do not attach importance to my gibberish.
Best regards,
Harry
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| Answer: |
Glenn Hawkins - 21/04/2012 22:13:39
| | | Hay guys! I am having a blast. What is this cool? You are such a sweet and no nonsense guy, Harry. For that mater, so is my pal Luis. I like you both very much. Luis you noted I complimented you. Also I didn't say it, but I have contemplated the same question also. Remember the paddle wheel in space? You just went on, as you say, a 'rant'. No problem. I like rants. It proves you have salt in your blood. Luis you have a pet peeve and I can respect that. Who needs free speech laws anyway. Peeve away. We all have opinions and you know what they say about opinions? Harry? They are like A holes, everybody has one. Luis come on now! We were getting along beautifully.
Harry, those times we all argued were awful. I wish I had shut up. I apologize to you both again. I was in error greatly, well socially for certain. Forgive my poor maters and I won't do it ever again-- but now I do like to banter about in a friendly way for a little bit, not often but now and then. Bless you all.
Glenn,
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| Answer: |
Sandy Kidd - 21/04/2012 22:54:26
| | | Luis, Glenn, Ravi, Harry and Shed Dwellers,
As Luis correctly stated and I would again like to reiterate at this point if I may that all my devices operate outside or below the point of saturation.
I have claimed that no usable, repeatable, differential can be gained from a system generating no angular momentum, well not that I can see at this time anyway.
Nitro’s device operated in this zone and produced positive results for reasons completely different from the operation of mine, but it is not easy to re-cock.
I am thinking that Harry has said that he still considers his “balance point” as an option.
I will say again that I have never managed to create a “balance point” in any of my experiments.
If the spinning wheel is in what is called precession in a mechanically accelerated system (or what I call saturation) there is no angular momentum present, which means the spinning wheel will accelerate upwards and inwards with nothing to slow it down or bar its progress.
Remember that as soon as the spinning wheel begins to accelerate upwards and inwards from its near horizontal position there is already no angular momentum left in the system.
Where accepted physics is wrong is in the fact that in a system rotating at a fixed speed there can be zero angular momentum or total angular momentum present depending on the rotation speed of the spinning wheel.
Also if the spinning wheel rotation speed is constant angular momentum can be likewise altered by altering the system rotation speed, but in a reverse fashion from accepted physics, the faster the system rotation the less the angular momentum.
I am not claiming Newton is wrong but physics certainly is.
Here is all the differential you will ever need.
Best regards,
Sandy.
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| Answer: |
Harry K. - 21/04/2012 23:02:13
| | | Hi Glenn,
Let's forget the past. I'm also not very proud about many comments I have posted here in this forum. At last we all try to increase our knowledge in gyro mechanics.
I like you too!
Best regards,
Harry
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| Answer: |
Harry K. - 22/04/2012 00:40:25
| | | Hello Sandy,
I read over your posting several times and I have a question.
You wrote:
"If the spinning wheel is in what is called precession in a mechanically accelerated system (or what I call saturation) there is no angular momentum present, which means the spinning wheel will accelerate upwards and inwards with nothing to slow it down or bar its progress."
Does this mean that what I call "forced precession", i.e. the precession of a gyro is not caused by gravity but by mechanically accelerated system is equal to what you call "saturation"? If so, than please excuse my ignorance because I always understood "saturation" as a zone where the angular momentum of the gyros dead weight mass overbalances the defelection (precession) force of the gyro. (I hope you understand what I mean.) This zone is limited from zero to a point, which I call "balance point" and beyond this point precession overcomes angular momentum of the gyro dead weight mass. Beyond this "balance" or "saturation" point there is no chance for angular momentum to overcome the precession (defelecting) force of the gyro. In this coherence I`m fully conform with your opinion. However, I know that I still owe a proof of this balance point by a test setup.
But if you claim that saturation occurs immediately in a mechanically accelerated system, which I understand as "forced precession", how has your prototype achieved inertial propulsion? Also Nitros setup works at last with forced precession caused by a rubber band, thus inertial propulsion should not have occur because angular momentum was not present to eliminate any force vectors.
So what do you mean that all your devices operate outside or below the point of saturation? Did they work under gravity conditions? I don't believe so. But at which conditions did they work?
Sorry again for my ignorance if you should have explained this matter elsewere in this forum.
Best regards,
Harry
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Glenn Hawkins - 22/04/2012 02:03:24
| | | How excellent. How excellent that all three of you are working in such advanced knowledge and theory. This is one of the best treads in at least one year.
Luis, you have no idea how lucky you are in life. My friend you could not guess how lucky you are. Good for you.
Glenn
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Luis Gonzalez - 22/04/2012 21:22:08
| | | "We have all said silly words, and I am not proud of mine either.
What we write reflects what's what we think. And what we think is what we are. If we read all that we have written, we may find out what we truly are."
Some types of comments help in developing seedling ideas into worthwhile concepts.
But other types of comments actually derail productive discussion of worthwhile ideas, causing lengthy distractions on irrelevant or less-important subjects.
Hi Harry,
We have once again found a point of disagreement. I hope it will be as productive as some of our previous discussions (without the personal attacks we are both guilty of, and hopefully without peripheral distractions).
(By the way, the limitations to observers is not determined by their LOCATIONS, but rather by the INFORMATION available to them. This is one of those peripheral or tangential points that may lead to much distraction, so I would prefer if we agree to discuss it only after we complete the main point, if we are still interested.)
- The main point is that the 3rd law DOES apply to some aspects of rotating systems (though you are correct that it may NOT apply to many angular-motion cases).
Let's explore.
The Torque and Counter-Torque of 2 mechanical components (e.g. a free-floating motor's Rotor and Stator) are in fact an Action and a Corresponding Reaction that Equal each-other in magnitude.
The above is in fact a basic display of the 3rd law as it applies to angular motion (even though they are angular action and equal-opposite reaction).
Please explain to me the reason that you would disagree with the last 2 sentences above Harry.
(Eventually we may get to discussing Sandy's "Saturation", Nitro's device, and the "Sweet Spot", but only if we manage to behave well.)
Best Regards,
Luis G
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Sandy Kidd - 22/04/2012 23:17:10
| | | Evening Harry,
I must apologise for any misunderstanding relating to saturation and “precession”
I will therefore endeavour to explain the whole thing to all, including readers new to this quest.
So Harry please do not feel offended if I tend to oversimplify my explanation but there are many individuals who will be coming into this cold as it were and probably need all the help they can get.
First of all my statements claims if you like are contrary to accepted beliefs relating to spinning discs under mechanical acceleration and it requires a large leap of faith to accept my word for it.
First time I saw this many years ago I sat in front of the device and repeated the experiment time and time again because I did not believe what I saw.
I suggest some simple experiments are carried out in order to prove the truth in what I say as no sensible person can be comfortable with what is about to follow.
About 30 years ago I built at a fair expense a device which I was convinced would deliver a lot of inertial thrust, but unlike my previous machine and to my great displeasure delivered no thrust at all.
However this device became the most valuable device I have ever built.
It taught me the truth about the behaviour of spinning discs when subjected to mechanical acceleration.
Let me explain.
A system in this context is of a disc mounted at one end of a shaft which has a fulcrum, swivel, whatever at the other end and whose purpose is not only to retain the shaft at that position but also allow the disc and shaft vertical freedom of movement.
In the interests of safety and in order to preserve balance in the system it would be wise to utilise the effect of a pair of horizontally opposed discs.
(Note I do not use gimbals just the raw disc).
A shaft is fitted to the system such that the discs and shafts can be rotated in a horizontal manner. We shall call this the main shaft as opposed to the shafts supporting the discs.
The system can be rotated and its rotation speed can be controlled from outside the system.
Similarly the rotation speeds of the pair of discs is synchronised and likewise can be controlled from outside the system.
Over the years I have discovered that the simplest and probably cheapest way of doing this is by utilising miniature (effectively 3 phase) brushless out-runner motors and associated electronic speed controllers (ESCs) which are very easy to control by radio.
In the case of the disc drives the power to the ESCs is through slip rings mounted at the bottom of the main shaft.
At this point the insertion of strain gauges to the experiment can be considered.
They should be mounted lengthwise on each of the disc supporting shafts and outputs fed to analytical instrumentation of your choice via other slip-rings.
OK we start to rotate the system and accelerate it up to say 250rpm.
I have found that this is a good starting speed
If strain gauges have been mounted note the readings.
This figure represents the maximum generation of angular momentum for this rotation speed.
Now accelerate the pair of discs to a rotation speed of say1000rpm.
The system will now be rotating at an angle several degrees below the horizontal.
It will now be noticed that there is a marked reduction in the reading from the strain gauges.
If we continue to increase the rotation speed of the discs the readings from the strain gauges will continue to reduce to a point where there is no angular momentum left in the system.
There is now no mass left to accelerate, consequently, there is nothing left to hold the discs down, but there is still a massive torque generated at the disc which drives the disc upwards and inwards.
This eliminates any possibility of a “Balance Point” Harry.
At exactly this point the discs will begin to accelerate upwards and inwards in an attempt to superimpose the axes of disc rotation over the axis of system rotation.
As you will gather this is the position of least action.
Gravity plays very little or no part in these proceedings, this is why I believe the term precession or forced precession in mechanically accelerated systems is in error.
If we return to the point where the discs begin their upwards and inwards acceleration it will be found that any increase in the application of disc rotation speed or any increase in the application of system rotation speed singly or together only serves to increase the upwards and inwards acceleration of the discs.
This is why I called this point the “Saturation Point”
In order to make a device deliver inertial thrust it is necessary to manipulate the angular momentum which is available in copious amounts below the saturation point.
Nobody said this was easy but it can be done.
For newcomers to this quest I recommend transposing centrifugal force for angular momentum when reading.
Definitely not the same thing but a trifle easier to understand for the ordinary guy, excuse me, person.
In the final analysis, whilst in saturation mode, and in light of the fact there is no angular momentum being generated, we are witnessing the spectacle of a pair of massless discs, being rotated at high speed without any acceleration.
This phenomenon applies from the point of saturation all the way to an elevation of 90 degrees, the energy level being the same at any point (angle) on the way.
The important thing is that angular momentum can be controlled by altering the disc rotation speed and /or the system rotation speed.
Provision is not made for this in the behaviour of rotating rotated masses.
I hope this helps to clarify my position.
In haste,
Regards,
Sandy.
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Glenn Hawkins - 23/04/2012 02:52:07
| | | Hi Luis,
“Can a device produce Linear Action with Opposite Reaction driving angular motion (i.e. rotation)?
Yes. Very sharp of you to come upon this. Harry's explanation is classic and scholarly and yes to him too. When all is said and done though, force weather applied to, or created from the straight or the circular order can be divided back into the other. In my onion this breaks the law of conservation of momentum. It appears that some actions in some possible ways do not fully result in equal reactions in the same venue of direction and place. It seems to me it should be difficult to argue that this division is not an exception to the order of physics.
Hi Harry,
You wrote, “. . . a rotating system cannot be an inertial system because inertial forces and thus fictitious forces occur in such a system.”
In mechanical reality can there be any force that is fictitious? I think the fictitious is in the fallacy of our designed method to use mathematics, particularly to do with centripetal, instead of stating the truth about mechanical action. I think in reality no action, or force cannot be fictitious, but erroneous human ideas can. The idea of centripetal is only a concept and not real. The idea is fictitious, not the misrepresented action. Action is true. What is your take on that? Could you be mistaken in your explanation of a rotation system not having inertia because it is closed?
You wrote, “Thus the 3rd law does not apply in your given example, independent from the location of an observer. Strictly speaking, there is no inertial system, because, so far we know, everything is in rotation.”
Right, Harry. In my limited ability I can agree. There are no straight paths or straight lines in the universe we know. Everything is either rotating or curving as in a partial rotation. All things curve, straight paths and even straight lines curve if they are long enough. Yet it is necessary that our math and mechanical laws rely on the straight line for thought and measurement, even to create the third law of motion and all similar laws. So strictly speaking, we must have an inertial system whether it is real or not, even though it is curving. Otherwise we can not explain, build or argue mechanical engineering. Do you you think your explanation is right? “Thus the 3rd law does not apply in your given example.” Because of curving action? If this is true, how could the same not apply to all other examples?
Hi Sandy,
How are you these cool days? Let us discuss your momentum findings, if this pleases you. I think the momentum reappears the instant the precessing wheel collides into a shock absorber. What do you think? I think the flywheel must be stopped quickly. If not the momentum (present but overwhelmed by deflections) will be absorbed as the wheel attempts to continue rotating, transferring the torque upwards. I have noticed the gyro with only one arm that you wrote about. When it touches down on the table it rolls around the pivot point as if breaks were being studding applied and they are by means of deflections. Of course the rolling is caused by the dead weight, and the breaks act by by the torque (in the place of momentum) being absorbed upward, but not enough to lift. But it seems that at the sudden collision of the wheel, the saturation is ended and the momentum of the rotating whee, that you could not measure, suddenly reappears with the force it should have had before deflections over powered it. I could try to postulated why these things are so.
You wrote, “The important thing is that angular momentum can be controlled by altering the disc rotation speed and /or the system rotation speed.”
Yes and also the hidden momentum can be controlled by controlling precession once it has accelerated to it's top speed from torque. Reaching that 'balance' from torque to precession, the wheel is acting like it is coasting with no opposite reaction at the controlled/pivot/hub area. If it is opposed while it is accelerating as in Nitro's experiment, you get opposite reaction. However the acceleration, when unopposed dose not cause opposite reaction as Harry and I once thought. When this is understood, propulsion is only a matter of designing to manipulating the resetting of mechanical oscillations based on coasting mass colliding with force into it's housing frame.
I may start a thread about this and everything there is to know about a gyro. After so many years I find the gyroscope is actually a very simple thing to know-- all of it.
Luis, “. . . without peripheral distractions.”! Shame on you. So you were out argued in your own created rant. Who cares? Come on. Stop it. Be a pall.
It is very nice to converse with you three intelligent creatures. You are human, Right? Earth people? Well greetings and so long temporarily from the Vega star system.
Glenn,
But you know my friends, I don't think anyone cares to know, for I never get a response from mechanical explanations of why and how things behave. Knowing how and why reveals an answer to every question. Harry is the only one who ever mentions deflections and yet understanding deflections revels everything-- everything there is to know about gyroscopes, except the exact magnitudes of the ratios involved,that is the math for best results. Yet knowing the mechanical is the only thing that really matters. With that you know all--everything. All else except the math that I do not know, but could follow. All else is sorter of cow fatter that we like to engage.
BE OF GOOD CHEER ALL.
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Glenn Hawkins - 23/04/2012 02:52:08
| | | Hi Luis,
“Can a device produce Linear Action with Opposite Reaction driving angular motion (i.e. rotation)?
Yes. Very sharp of you to come upon this. Harry's explanation is classic and scholarly and yes to him too. When all is said and done though, force weather applied to, or created from the straight or the circular order can be divided back into the other. In my onion this breaks the law of conservation of momentum. It appears that some actions in some possible ways do not fully result in equal reactions in the same venue of direction and place. It seems to me it should be difficult to argue that this division is not an exception to the order of physics.
Hi Harry,
You wrote, “. . . a rotating system cannot be an inertial system because inertial forces and thus fictitious forces occur in such a system.”
In mechanical reality can there be any force that is fictitious? I think the fictitious is in the fallacy of our designed method to use mathematics, particularly to do with centripetal, instead of stating the truth about mechanical action. I think in reality no action, or force cannot be fictitious, but erroneous human ideas can. The idea of centripetal is only a concept and not real. The idea is fictitious, not the misrepresented action. Action is true. What is your take on that? Could you be mistaken in your explanation of a rotation system not having inertia because it is closed?
You wrote, “Thus the 3rd law does not apply in your given example, independent from the location of an observer. Strictly speaking, there is no inertial system, because, so far we know, everything is in rotation.”
Right, Harry. In my limited ability I can agree. There are no straight paths or straight lines in the universe we know. Everything is either rotating or curving as in a partial rotation. All things curve, straight paths and even straight lines curve if they are long enough. Yet it is necessary that our math and mechanical laws rely on the straight line for thought and measurement, even to create the third law of motion and all similar laws. So strictly speaking, we must have an inertial system whether it is real or not, even though it is curving. Otherwise we can not explain, build or argue mechanical engineering. Do you you think your explanation is right? “Thus the 3rd law does not apply in your given example.” Because of curving action? If this is true, how could the same not apply to all other examples?
Hi Sandy,
How are you these cool days? Let us discuss your momentum findings, if this pleases you. I think the momentum reappears the instant the precessing wheel collides into a shock absorber. What do you think? I think the flywheel must be stopped quickly. If not the momentum (present but overwhelmed by deflections) will be absorbed as the wheel attempts to continue rotating, transferring the torque upwards. I have noticed the gyro with only one arm that you wrote about. When it touches down on the table it rolls around the pivot point as if breaks were being studding applied and they are by means of deflections. Of course the rolling is caused by the dead weight, and the breaks act by by the torque (in the place of momentum) being absorbed upward, but not enough to lift. But it seems that at the sudden collision of the wheel, the saturation is ended and the momentum of the rotating whee, that you could not measure, suddenly reappears with the force it should have had before deflections over powered it. I could try to postulated why these things are so.
You wrote, “The important thing is that angular momentum can be controlled by altering the disc rotation speed and /or the system rotation speed.”
Yes and also the hidden momentum can be controlled by controlling precession once it has accelerated to it's top speed from torque. Reaching that 'balance' from torque to precession, the wheel is acting like it is coasting with no opposite reaction at the controlled/pivot/hub area. If it is opposed while it is accelerating as in Nitro's experiment, you get opposite reaction. However the acceleration, when unopposed dose not cause opposite reaction as Harry and I once thought. When this is understood, propulsion is only a matter of designing to manipulating the resetting of mechanical oscillations based on coasting mass colliding with force into it's housing frame.
I may start a thread about this and everything there is to know about a gyro. After so many years I find the gyroscope is actually a very simple thing to know-- all of it.
Luis, “. . . without peripheral distractions.”! Shame on you. So you were out argued in your own created rant. Who cares? Come on. Stop it. Be a pall.
It is very nice to converse with you three intelligent creatures. You are human, Right? Earth people? Well greetings and so long temporarily from the Vega star system.
Glenn,
But you know my friends, I don't think anyone cares to know, for I never get a response from mechanical explanations of why and how things behave. Knowing how and why reveals an answer to every question. Harry is the only one who ever mentions deflections and yet understanding deflections revels everything-- everything there is to know about gyroscopes, except the exact magnitudes of the ratios involved,that is the math for best results. Yet knowing the mechanical is the only thing that really matters. With that you know all--everything. All else except the math that I do not know, but could follow. All else is sorter of cow fatter that we like to engage.
BE OF GOOD CHEER ALL.
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Glenn Hawkins - 23/04/2012 17:26:27
| | | I was wrong, Harry. I quote a report.
"The Law of Conservation of Momentum is much more powerful than Newton's Second Law. No experiment to date has shown a violation of the Law of Conservation of Momentum. It applies to objects moving very fast or to objects inside the atom - both places where Newton's Second Law fails."
Newton's second law of motion states that force is equal to that object's mass multiplied by its acceleration, or f=ma.
I guess you are happy now, Sandy. Your nightmare that there are errors in organized physics is now confirmed by mathematicians. Sleep well tonight. :-).
Glenn,
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Harry K. - 23/04/2012 19:35:15
| | | Hello Luis, Hello Glenn,
I thought over what I wrote and noticed that I made a big mistake (did I mention that Im not an expert in this matter?).
An inertial system -assumed it could be really present - allows angular rotation, momentum or whatever, for instance two counter acting forrces with a parallel distance to each other is identical to a torque which will cause rotation or rather torque. The operative point is if the frame of reference remain at rest or not.
Your good objection regarding acting and counter acting of the stator and rotor of a motor is correct, however the acting and counter acting rotations are WITHIN the frame of reference. This means that the frame of reference itself will remain at rest:
- the rotor is trying to rotate the system in one direction
- the stato is trying to otate the system in counter direction
That means that both actions will cancel out each other and thus the frame of reference will remain at rest.
On the other hand If an observer is LOCATED e.g. in the frame of reference of the stators plane, he would notice a force which try to move him in a straight line linear outwards away from the rotor which rotates at double speed as it do in reality. The observer does not have more INFORMATION what really happens. Because his frame of reference is in rotation and thus creates fictitious forces (centrifugal force in this case) the system cannot be an inertial system where Newtons laws are valid.
Now it's easier (for me) to understand what really happens with your stated example at the beiginning of you thread:
"EXAMPLE:
1 - A spoke on a flywheel uses a strong spring (or a small explosion) to fire off a mass in a direction TANGENT to the flywheel's perimeter.
2 - The ejected mass travels in a straight Linear trajectory.
3 - The Reaction force accelerates the mass of the device into an Opposite Linear direction but most important it also accelerates the mass of the flywheel into Angular Rotation."
Part one of No. 3 is confirmed.I assume the flywheel isn`t in initial rotation?
I see part two a little bit different:
The strong spring or small explosion will on the one hand accelerate the mass in one direction and the complete flywheel wobbling in the counter direction. Note that the flywheel will not spin up because only one force, the reaction force, will act at the flywheel, but to spin up the flywheel, a counter force in addition with parallel distance is needed to cause the necessary torque. However the flywheel will wobble straight in counter direction to the fired mass because the counter force is acting with a distance to the center of the flywheels mass and thus the inertia of the flywhhel delivers a part of counteracting force to cause a torque. I hope this is understandable with my very special English diction.
Anyway, the sum of all acting forces and torques will cancel out each other and thus the frame of reference will remain at rest.
Best regards,
Harry
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Harry K. - 23/04/2012 19:45:59
| | | Hello Glenn,
Thank you for your comments. I`m always interested in your opinion. Please apologize if I did not answer sometimes in the past. You are one of the most experienced guy regarding gyroscope issues here in the forum and thus I very welcome your opinon and comments.
Thanks!
Harry
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Harry K. - 23/04/2012 20:29:41
| | | Hello Sandy,
Thank you very, very much for your clarifications! I`m ashamed that you have to repeat again and again your saturation point/zone to me.
One section in your explanation is very intersting for me:
"Now accelerate the pair of discs to a rotation speed of say1000rpm.
The system will now be rotating at an angle several degrees below the horizontal.
It will now be noticed that there is a marked reduction in the reading from the strain gauges."
Does this mean that at a special rotation speed a constant angle above horizontal plane appears? Is this the situation, the system rotatates with a fixed angle about horizontal plane, what you call saturation point and beyond this point an increase of system speed or rotation speed of the gyros will initiate the upward and inward movement of both gyros?
If above is correct, than I understand your "saturation point" as my called "balance point", it would have exactly the same effect, a balance between centrifugal forces of the gyros dead weight mass and the torque caused by forced precession.
The strain gauges indicate that the centrifugal force decreases and disappears during upward and inward movement and hence angular momentum decreases and disappers at well. I entirely agree with this. Centrifugal force is acting against precession forces and the smaller the rotation radius of the flywheels, the smaller becomes the centrifugal force and so it cannot resist the precession force (torque) which is comparatively bigger, depending of the rotation speed of the flywheels.
Did you noticed any propulsion effect if the the gyro hub was rotating with an fixed (if so) angle above horizontal plane?
Thank you again for your excellent explanation and your patience!
Best regards,
Harry
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Sandy Kidd - 23/04/2012 22:40:52
| | | Hello again Harry,
More saturation stuff.
I did not think I could get away that easily.
What I was trying to describe was a very typical mechanically accelerated system.
Providing the discs are not rotating, it does not really matter if the system rotation speed is 250rpm or 2,500rpm the discs can never reach a truly horizontal position because of gravitational effects.
The insertion of the1000rpm disc rotation speed was a speed I selected as probably not capable of sending the disc into saturation but this depends purely on the diameter and quality of the disc and the geometry of the system, but mainly the radius it is being accelerated through.
I could have been a bit more precise or specific with my values, at least I could have enlarged on what I was intending.
Accept my apologies for that, Harry.
I was trying to direct the reader’s attention to the fact that as long as the system remains out of saturation the discs will remain near but below the horizontal.
The point of interest here is that without instrumentation or results from prior experiments the quantity of lost angular momentum is unknown and invisible until the system reaches the saturation point where the action becomes visible.
As soon as the discs reach the horizon the system is already in saturation mode.
It took me sometime to realise this, as I repeatedly started a device which had immediately accelerated into saturation and I could not for the life of me figure out what was going on.
The point is that we now know that it is variable and there can be the total angular momentum generated with the discs not rotating i.e. a dead weight, or there can be zero angular momentum generated at the saturation point.
That gives us a fair range in which to produce our differential however we intend to do it.
Saturation is really a pain in the rear parts, but we must beware of its existence. What is important is in knowing that controllable changes can be made to angular momentum and dare I say it, centrifugal force.
Is this any help, Harry, sorry still no balance point.
Nice if there was.
Regards,
Sandy.
PS
It requires a little bit more than this to generate any propulsive effects, but we need to go this way and it’s a start.
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Luis Gonzalez - 24/04/2012 01:59:09
| | | Harry,
Your basic answer, stating that the example I presented "does NOT break the 3rd Law", is correct.
That is the main point that the example is intended to drive home (i.e. mixed interaction among linear and angular motions do NOT necessarily break the 3rd Law). Thank you for answering honestly.
In the examples presented, the sum of all the actions and reactions equal zero, which is important to preserving the 3rd Law but I would Not say that "things remain at Rest". I guess that depends on the perspective / frame of reference (this is one of those things I would rather not pursue at this point, as it distracts my intended goal in this thread - we can discuss later or elsewhere).
Finally, the original example I presented does NOT depict a CLOSED system so preserving the 1st Law (Conservation of Linear Momentum) is not relevant; after all the example presented uses a propellant (spring or explosion) to EJECT a mass.
Harry and Sandy,
Thanks for clarifying the comments about "gyros rotating below the horizontal" (your clarification was a big relief) .
I am also glad Sandy that you are helping Harry to refreshing his memory regarding the Saturation Point.
We have previously discussed the dynamics of Saturation, though in a slightly different context.
I was rather hoping we could discuss how Nitro's device produces mass displacement from inside the Saturation Zone".
Also, by now you have probably guessed that I believe the 3rd Law is NOT violated in Nitro's experiment (I also believe the 3rd Law remains intact even inside the Saturation Zone). That is one reason that I presented the original example, to set up the foundation to my argument.
Further, I believe that proving gyro propulsion does NOT negate or challenge the laws of motion, will go a long way toward acceptance by some segments of academia. Perhaps that's just an opinion but I believe it.
Glenn,
Don't worry, the path to each success is marked by many failures.
A little humility can help a lot. Hang in there.
Perhaps you can take a fresh approach at exceptions to the laws of physics.
Best Regards,
Luis G
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Luis Gonzalez - 24/04/2012 02:04:54
| | | Hi All,
I will not be responding too quickly, as I am going back to work after my sick day today.
Regards,
Luis G
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Momentus - 26/04/2012 11:38:58
| | | Hi Luis
In your example, the flywheel has a centre of mass offset from the point of action.
When I did my experiments with this configuration it was easier to strike the ‘spoke’ with a moving mass than to eject it. The result is the same, the flywheel both rotates and translates.
Remember that momentum is a quantity of velocity, whereas energy is a quantity of velocity squared. Assume that the flywheel has the same mass as the impacting object, then by momentum addition they will continue in a right line at half the speed, twice the mass and momentum is conserved.
Half the square of the speed however leaves energy to be conserved. This is absorbed by the rotation of the flywheel.
In your example the two masses will move apart with the flywheel rotating.
Like Harry K said energy is conserved. Like Glen’s pen moves away spinning as it goes. (translates and rotates)
No way to use that one as a propulsion device so back to the drawing board.
Momentus.
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Momentus - 26/04/2012 12:13:06
| | | Sandy,
A eureka moment (for me), finally translating “saturation” into my own world view.
I posted a method of creating angular momentum (id 460) where the spin axis and the precession axis are interchanged.
This is the same as the saturation point, when the disk speed/momentum equals the system speed/momentum.
When the disk speed is zero, the system axis is the spin axis and to increase the system speed you need torque. When the disks start spinning, they are in effect precessing, no torque needed to spin them up. When they spin fast enough the spin axis flips to the disks and effects most untoward and unexpected appear. Centripetal forces vanish.
Food for thought there.
Momentus
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Luis Gonzalez - 27/04/2012 02:24:16
| | | Momentus,
Why must I be misunderstood.
Am I so unclear or are my words not read well enough?
Any thinking man knows that the device in the example is NOT INTENDED to produce propulsion, much less inertial propulsion (after all it EJECTS a mass. Right?).
It's a no-brainer.
Did you miss my real reason for the example presented Momentus?
It is solely intended to prove that the 3rd Law is not necessarily compromised by mixing of linear and angular forces.
Do you understand why I am trying to prove that point?
Regards,
Luis G
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Luis Gonzalez - 27/04/2012 02:33:15
| | | Sorry Momentus,
I just read your posting to Sandy and that drove the point home for me.
It appears we perceive the world from completely different perspectives.
That is a good thing. We would not want the whole world thinking exactly the same way.
Regards,
Luis G
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Luis Gonzalez - 28/04/2012 00:15:53
| | | Momentus,
I now see that my intended proof is on its way down the drain.
In short, the much discussed "external observer" will see an equal and opposite reaction occurring.
I will write a more complete explanation for anyone who did not catch the implications in Momentus' posting to me.
Regards,
Luis G
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Luis Gonzalez - 28/04/2012 04:26:31
| | | Hi Momentus,
My sincere apology for focusing my immediate response on the less relevant comments.
Hi Everyone else,
I often have difficulty deciphering what Momentus writes regarding physics.
However in this case it compelled me to think further on the initial example in this thread.
I compared variations between the mass of the EJECTED object as it compares to the mass of the flywheel etc.
And then I realized that BOTH of the Straight-Line Opposing motions (in the main example) would be affected by the partial conversion of force to angular momentum.
That is BOTH opposing linear motions would be SLOWER as compared to an equivalent experiment if none of the momentum becomes angular (i.e. NO flywheel rotation).
In other words, the ejected mass and the device's opposing straight-line motions become SLOWER because part of the spring's "work" was used for turning the flywheel into angular motion.
The most interesting part is that BOTH opposing linear motions remain EQUAL to each other in velocity! This is what really happens in the main example (not what I originally thought).
The external observer sees the action and reaction as equal and opposite...i.e. normal.
I suspect this is what the clever Momentus in fact stated, in such concise manner that I found difficult to consciously decipher. (I truly don't expect he will confirm or deny...)
The implication is that the main example presented, in fact FAILS to Prove that equal and opposite motion conversions between linear and angular motion do necessarily preserve the 3rd Law.
The example does not Prove or Disprove anything we did not already know.
It is a disappointment, as my initial conclusion was an error.
All relevant thoughts are welcome.
Best Regards,
Luis G
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Glenn Hawkins - 28/04/2012 06:09:44
| | | Hi Luis,
You have it right. You and I are the two fast draws on here. It isn't often, but we have pulled trigger too fast a few times, but later through reflection realize the right answer and get it right always. I think that is just fine.
Another ditty: A rubber ball strikes the flat surface of a more massive object in space and bounces backwards. It’s easy isn’t it? The object both yields (slowly) and pushes back, and just as in your example momentum is also divided.
I think Momentum forgives us sometimes. So just as Harry advises, keep a cool tool. Till later. . .
Best Regards Glenn,
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Luis Gonzalez - 28/04/2012 18:17:50
| | | Deal All,
By now we should all have wrapped our mind around what happened in the main example, up to a point.
However, there is still one main lose-end to resolve.
Where is the COUNTER-Torque to the Torque that creates the angular motion?
Perhaps a better question may be:
Where does the sneaky COUNTER-Torque reveal itself?
My normal Inclination (pardon the pun) makes me suspect that the TRUE Direction of the 2 Resulting Straight-Line motions will bear the effects of the seemingly Missing COUNTER-Torque.
Again, this curt answer leads to yet another question (how prosaic...!!!).
(Sorry to bore you all but I must seek conclusion .)
Q. How are the 2 resulting Straight-Line motions affected by the COUNTER-Torque?
Do they remain perfectly aligned?
Or is a slight angle of deflection introduced to ONE of the 2 motions?
If so, Which one of the lines is deflected (the ejected object, or the crude device)?
I suspect that both Straight-Line motions will be proportionally DEFLECTED by equal angles so that they remain in perfect line with each other.
If anyone knows the actual answer, I welcome it.
I am sorry to say it, but this subject is probably as dead as dead can be.
So, the 3rd Law is still up in the air.
Regards to All,
Luis G
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Glenn Hawkins - 28/04/2012 22:32:31
| | | Surely everyone knows the answers by now including yourself or course.
I think you are right in saying this subject is probably as dead as dead can be. I have succeeded in reducing the gyroscope to a few simple functions and there is nothing more mechanically to know. Only the mathematical ratios are left to learn, but I sometimes think no one wants to know. I will respond to you sometimes. Otherwise I guess the actual solutions are valuable to me alone.
Regards Glenn,
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Luis Gonzalez - 28/04/2012 23:30:17
| | | I only meant that the subject in this thread's example is dead (not inertial propulsion or gyro dynamics)!!
There are years of learning about spin dynamics ahead.
Regards,
Luis G
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Glenn Hawkins - 29/04/2012 00:53:40
| | | No way.
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Harry K. - 29/04/2012 11:39:56
| | | Hello Sandy,
Sorry for the long delay in answering.
Thank you again for your clarification! I agree with you regarding the behavior of a gyro system in saturation, however, I'm still convienced that there must be a situation of balance between angular momentum caused by the gyro's dead weight mass and forced precession torque. I think it's difficult to uncover this point of balance if you do not search for it, i.e. you have to choose suitable parameters to find it.
Anyway, I know that I only may convince you with a prooved setup. I hope to find enough time to do this experiment in near future.
Thanks again for your clear explanations!
Best regards,
Harry
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Harry K. - 29/04/2012 11:51:12
| | | Hello Luis,
Regarding your last post I hope to get you correctly.
In your given example the wobbling (=angular momentum) and linear moving flywheel (inertial momentum) are the COUNTER reactions by the action of the spring or small explosion.
Again, the force caused by the spring or small explsion with its distance to the flywheel's center of mass causes the angular momentum (wobbling) as well as its the linear movement.
Or didn`t I get you correctly?
Best regards,
Harry
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Luis Gonzalez - 30/04/2012 01:35:59
| | | Hi Glenn,
Ok smart guy, why does the sweet-spot occur in Nitro's device (in the video)?
No insult intended; just trying to verify if you really know all about gyros as you stated.
Best regards,
Luis G
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Luis Gonzalez - 30/04/2012 01:42:44
| | | Hi Harry,
Thank you for your effort and comments. I think you understood the original example just fine.
Since then, the example has been proven to have errors, which make it not usable for the intended proof.
I think it's probably better to spend time in more productive threads.
Thanks again and Best Regards,
Luis G
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Glenn Hawkins - 30/04/2012 02:31:19
| | | The sweet spot is the perfect area to strike a golf ball for maximum good effect, a game invented in Sandy Kid's back yard I believe. Ask him.
I know. Who gives a skittt? I would not have used the term to related to the exact ratio of magnitudes between angular momentum, right angle hub acceleration, speed and mass of an object in the path of precession being moved. I applaud Nitro highly and the experiment, but I cannot be wrong not even by his reckoning, because nobody knows what the hell I am talking about and I don’t care.
I said all the mechanical issues of a primary flywheel being force to rotate at a right angle around a secondary point of reference is understood.
Bless your heart, Luis.I know your are a nice person and both committed and smart. I know you will be good to your family. Nothing else really matters.
Glenn,
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Glenn Hawkins - 30/04/2012 02:33:05
| | | 'Kidd' Sorry Sandy.
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Harry K. - 30/04/2012 11:21:53
| | | Hello Luis,
Your questions were good and thus this thread made sense because we all can always learn new insights about this subject.
Best regards,
Harry
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Glenn Hawkins - 30/04/2012 14:05:54
| | | I am sorry Luis,
Too much cool-aid last night. I see I needed a proof reader everywhere I posted.
Nitro doesn’t want insights from other people into his whys and hows, so I have stayed away. But I will give you my understanding of his invented term, a term that I would not have coined. Nitro’s sweet-spot is a balance, a ratio between opposing forces. These are, angular momentum verses right angle torque. When they are acting in a certain balance, I believe Sandy calls this saturation, which is a condition as I understand him where deflections have become stronger than centrifuge.
As you and everyone clearly knows, Nitro’s first action reaction is relatively strong. The hanging weight and frame assembly would continually separated if they were not fastened together. They act a lot faster than the later oscillating pendulum effect upon the total apparatus, which moves it only about ½” and ceases. The small distance and stop indicate the condition is actually mass movement. I did not see or realize any collision-like effects occurring, which might have resulted in actual thrust.
Why the balance difference works, is the better question, than what is the sweet spot.
If angular momentum were stronger, precession would be slower and weaker.
If right angle torque were stronger, precession would take on the properties of rotation.
The best balance between the two, the sweet spot, which is mathematical, will be the best conducive to Nitro’s experiment.
Sandy Kidd has danced into and around hands-on all the way up and down these conditions until he understands backwards and forwards, but he understands them the Kidd way, which is not necessarily the Nitro way.
When I was in the military I learned there were three ways to do anything, the right way, the wrong way and the Army way. Choose your way. Sandy and Nitro did.
Your answer: The sweet spot is a balance between two opposing forces that best allows a desired effect. In this case a movement.
Happy trails to you and Dale and Trigger, until we meet again,
Glenn
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Luis Gonzalez - 06/05/2012 05:21:53
| | | Hi Blaze,
Do me a favor and post a brief statement about the 3rd Law, in this thread.
It can be as simple as you wish.
Best Regards,
Luis G
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Blaze - 06/05/2012 05:44:29
| | | Newtons 3rd law:
The mutual forces of action and reaction between two bodies are equal, opposite and collinear.
Collinear means in the same straight line.
The 3rd law is often stated as: for every action there is an equal and opposite reaction.
Blaze
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Blaze - 06/05/2012 05:57:53
| | | A rather more involved explanation of Newton's 3rd Law:
A force is a push or a pull upon an object that results from its interaction with another object. Forces result from interactions! Some forces result from contact interactions (normal, frictional, tensional, and applied forces are examples of contact forces) and other forces are the result of action-at-a-distance interactions (gravitational, electrical, and magnetic forces). According to Newton, whenever objects A and B interact with each other, they exert forces upon each other. When you sit in your chair, your body exerts a downward force on the chair and the chair exerts an upward force on your body. There are two forces resulting from this interaction - a force on the chair and a force on your body. These two forces are called action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is:
For every action, there is an equal and opposite reaction.
The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.
Blaze
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