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3 May 2024 13:18
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Question |
| Asked by: |
Luis Gonzalez |
| Subject: |
An answer to “Does spin-deflection (precession) stop immediately when torque is removed?” |
| Question: |
This forum is a great place to hide esoteric knowledge about gyros. The correct answer, to the question asked in this thread, is explained openly in this forum. However, the answer may not be easily found, among the abundance of irrelevant, erroneous and misleading information; bits of wisdom are hidden in a sea of misunderstandings, confusions, and ranting.
So… why does the motion of spin-deflection (precession) appear to stop immediately, without apparent need to slowdown, when torque becomes absent?
Luis Gonzalez |
| Date: |
3 March 2008
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Answers (Ordered by Date)
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| Answer: |
Luis Gonzalez - 03/03/2008 15:05:05
| | | The explanation, as to why spin-deflection (precession) looks like it stops without need to slow down is as follows:
When a force is used to change the direction of a moving object, the resulting deflection occurs only while the force is applied, and not a moment longer. Visualize an object traveling in space (e.g. a space ship). A lateral force changes the object’s direction only while the force is applied; as soon as the lateral force is stopped, the object stops changing direction and its motion returns to a straight line again.
In other words, there is a difference between 1) changing the direction of an object’s motion, Vs 2) changing the speed of an object’s motion. The first one has a dynamic effect only as long as the force is applied, while the second permanently affects the magnitude of the motion. Changing the speed of an object affects its momentum, but changing the direction of an object does not affect its momentum (only its direction). And if the original motion was spin, then the spin just continues, but the reorientation motion stops immediately.
The explanations above may appear obvious, but they also help to explain the question posed in this thread.
This explanation addresses the reorientation of the flywheel during spin-deflection/precession because that is the motion that is being changed (deflected) by the torque. However, the “Displacement” created during spin-deflection (precession) does indeed have momentum, even though many of the best minds in this forum have denied it, until recently (I even started to doubt it myself, at one point). In other words, “Displacement” has momentum even though “Reorientation” does not. “Reorientation” stops immediately” but “Displacement” does not.
We know that the displacement must accelerate from zero to its maximum velocity of spin-deflection (precession) and must also decelerate to get back to zero velocity. Harry K., Momentus and I have explained how that works (though it can take time to sink-in at the beginning). However, the deceleration (and acceleration) occurs while the flywheel maintains a spin-velocity (and momentum), which is many times grater than the velocity of spin-deflection (precession). Average gyros can spin over 200 times faster than their rate of precession; therefore, precession contributes less than one-half of one percent (1/2%) of the gyro’s net momentum, which is easily lost in the overall dynamics. The acceleration and deceleration of spin-deflection (precession) occur within a brief moment and their effects compete against a combination of all types of friction (e.g. spin, pivot, air, and other types of friction), plus the effects of all deadweight involved.
It’s no wonder that we have difficulty demonstrating experimentally what textbooks have cryptically stated; I must admit it’s not easy to find English textbooks that explain these concepts clearly if at all. Most physics books provide the basics and we need to develop and derive the intricacies ourselves (think back to your college days).
In a previous thread, I listed a number of components with which to perform non-intrusive gyro experiments designed to test the effects of displacement’s acceleration and deceleration. The experiment involves clearly marking strategically placed points on the gyro’s components, with white or fluorescent paint. The gyro should be setup to start inside a dark room where a fast strobe-light can be adjusted to specific frequencies. First the gyro is spun, lights are turned off, a cheep camera with long exposure is clicked on, the strobe light is turned on, and then the gyro is dropped. One can also drop a golf ball simultaneously to determine the timing via the strobe-images (it would not hurt to have a stopwatch as well).
The most expensive part of the experiment is the strobe-light at about $300. The rest can be purchased for peanuts, and a closet is sufficient space for the experiment. I suggest using an extra-long axle to place upon the tower because the motions will be magnified by the length of the precession radius.
Correct answers to difficult questions are most often based on known concepts, which are relatively easy to understand. Rarely are the correct answers based on equations that have nothing to do, or have nothing in common with the equations of established science. For example, Einstein’s revolutionary motion equations were built and expanded upon, Newton’s motion equations.
Some people say that gyro-propulsion does not need deep complex analysis in this forum. Yet the same people claim that the correct mathematics does not yet exist for the relevant gyro phenomena! These two statements are contradictory and must result from disjointed thinking, but I’m not sure whether it is intentional or not.
I know that digging into the details of a complex subject eventually yields simplicity, even though the journey itself is not as easy. The only way to connect the math with the phenomena is through in-depth analysis of the experimental results (the analysis should be grounded on what is already known).
As the reorientation of spin-deflection (precession) comes to an immediate stop, because the torque is removed, the displacement velocity must decelerate causing a short-lived spin-deflection at 90o to the deceleration “force”.
Note, as the reorientation of spin-deflection (precession) stops, the displacement’s momentum creates a force in the same direction of the original precession (similar to placing the brakes on your car), which causes a new spin-deflection (precession) in a new 90o direction.
In conclusion, the “Reorientation” of spin-deflection (precession) does come to an immediate stop. However, the “Displacement” causes a short-lived new spin-deflection, whose new direction may be interpreted as a 90o change in direction of the original spin-deflection.
Thank you,
Luis Gonzalez
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| Answer: |
Harry K. - 03/03/2008 23:16:25
| | | Dear Luis,
Thank you for your well sophisticated summary! However, I can not agree entirely.
But first I have to know what you exactly understand with "displacement" in this context. Do you mean the precession movement of an overhung gyro, i.e. the precession movement beyond the center of spinning mass? - If so the term "displacement" is not correct in my opinion, because there is no resulting mass displacement of an overhung gyro after one complete precession circulation.
The terms should be chosen well to avoid misunderstandings and confusions in the future. ;-)
Thanks!
Harry
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| Answer: |
Luis Gonzalez - 04/03/2008 01:24:18
| | | Dear Harry,
You are correct; I am using the word “displacement” to mean the motion that changes the center of the spin object in precession (as well as in mechanically induced spin-deflection).
Why don’t you suggest a better concise term (not too wordy please).
The fact that a complete precession circuit returns the mass to its original position (more or less) does not mean that the motion did not take place. A tune can start and end with the same note and still takes place.
I am looking forward to your naming choice.
Best Regards,
Luis Gonzale
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| Answer: |
Harry K. - 04/03/2008 19:32:41
| | | Dear Luis,
Thank you for your confirmation. Your comments regarding a complete precession circuit are certainly correct, but the use of term "displacement" may wrongly be interpreted in this forum as "moving mass" (=propulsion!). That's the reason for my objection.
I could live with term "displacement" but I won't use it. I will use the paraphrase "precession movement around center of spinning mass" and "precession movement beyond center of spinning mass". I know it's along phrase, but in fact correct and not misleading.
Now I will come back to your thoughts and statements.
Please note that all of may comments in the other thread are related to a gyro system which will precess around its center of spinning mass ( no displacement! ;-) ), i.e. not a overhung gyro system.
You wrote:
"When a force is used to change the direction of a moving object, the resulting deflection occurs only while the force is applied, and not a moment longer. Visualize an object traveling in space (e.g. a space ship). A lateral force changes the object’s direction only while the force is applied; as soon as the lateral force is stopped, the object stops changing direction and its motion returns to a straight line again."
I think this imagination is made too easy and not very helpful to explain the accordant gyro behavior. In your example there is one force acting to change the direction of one moving mass object. A gyro, however, consist of endless moving mass points around its circumference but the tilting force (torque) affects only to one mass point at 90 deg. and a limited amount of mass points between 0 and 180 deg. of the applied tilting force. That's a big difference!
Also I'm sure that the velocity of the space ship in you example will be íncreased. Note, that the new direction is caused by the resulting of 2 velocitiy vectors, the formerly velocity vector of the space ship and the velocity vector caused by the applied 90 deg. force.
So you see there is no difference in your example, because the speed of the space ship will be increased. And therefore your further statements are not correct as well.
Furthermore in contrary to your opinion I'm convinced that momentum will be stored in precession movement around center of spinning mass. The amount of momentum however will be very small in relation to the momentum of spinning mass but it is present.
For better understanding imagine a gyro model consisting of only 2 diametral spinning masses. Now apply a tilting torque exactly at the point of time when the mass velocity vector will be 90 deg. to the tilting torque. The spin axis will be deflected for a short time in the same way as a "normal" gyro would do. In this case there is no "dead mass" present, which have to be accelerated in precession direction. In only this case your stated assumption would be correct.
Now imagine a gyro with 4 masses even placed around the circumference of the gyro. Now repeat the procedure and apply again a tilting torque at that time when the mass velocity is 90 deg. to a tilting torque. In this case the precession movement must accelerate the 2 "dead masses" which are not under influence of the tilting torque.
Do you recognize the difference? A gyro with endless spinning masspoints will have the same behavior. There is a intersection of mass points which are under influence of the tilting torque and a intersection of mass points which are under influence of precession movement and therefore these mass points are under influence of mass inertia as well.
So far to your statements. It would be nice to read your or any others comments.
Best regards,
Harry
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| Answer: |
EDH - 09/03/2008 08:42:12
| | | “Does spin-deflection (precession) stop immediately when torque is removed?”
Yes. Part of the magic of a gyro, is that all forces compute simultaneously. This is one reason why the calculations for a precessing gyro can be incredibly complex. No wonder I prefer to use a computer! So, yes, any change in the gyros torque or position will be simultaneously translated in to the expected motion.
EDH
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| Answer: |
Luis Gonzalez - 10/03/2008 20:18:46
| | | Dear Harry,
When it comes to conventional gyro physics, you are probably one of the best technically trained individuals in this forum. However, I believe we all share a common interest and tat is to expand common gyro wisdom, as we understand it (some better than others), and I think you are share that common interest too.
Despite different points of view, I hope we can learn new things from each other that are useful to our objectives. I also think that we may both have a couple of errors in our thinking (especially in areas that we have not spent sufficient time thinking about (it’s a big subject)).
I sincerely hope we can both correct our errors, but frankly I am more interested in finding solutions to my own errors.
From your posting, it looks like you perceive the torque (applied to tilt a spinning flywheel rim) as if it drags portions of the flywheel in a manner similar to dragging deadweight (at least to some degree). By this reckoning, you expect the deadweight to acquire momentum as it is dragged “around the center of spinning mass” by precession (not including “beyond center of spin mass”). I can see the reason for your point of view about this interaction; however it’s missing one thing; this point of view underplays the fact that we are not moving deadweight but rather changing the direction of an existing motion (nothing more).
Let’s take a closer look at the rocket ship example; first compare the ship to individual point-masses (NOT to the whole gyro), then consider the following, which is most important:
The velocity of the spaceship in my example can not increase when a constant acceleration is applied at 90o to the motion of the ship (without deviation). In fact, the ship will end up in a circular path as its direction changes constantly because the net force behaves exactly as Centripetal force does, therefore no change in velocity, only change in direction. Please conduct the necessary experiments etc, to visualize the dynamics.
What if the applied force results from an acceleration that is greater then the square of the velocity (of the ship) divided by the radius of the circular path. (A > V**2 /R) In such case the radius will become smaller until the object (rocket) and configuration does not permit the radius to shrink any more. Only at that point will the object begin to spin faster around its center of mass at a speed that exceeds its original speed (as long as the force is applied at 90o of the motion). Note that as long as the velocity does not increase, the momentum does not increase either.
If my rocket ship example is accurate when compared to a point-mass, then it should be relatively clear that as each point-mass is deflected, from its established angular (spin) motion, its interaction is the same (close enough) as the deflected rocket.
In one of my previous writings, I argued that the velocity of the flywheel may increase as it is augmented by the rate of precession “around the center of spinning mass” (not including motion “beyond center of spin mass”). However, we know that the increase velocity can not occur on all points of the rim at the same time because the orthogonal positions do not increase in velocity (around the center of spinning mass). The spin will increase and decrease ever-so-slightly every quarter turn of the spin.
If there are any extraordinary phenomena (such as propulsion or energy) that can be obtained from spinning objects, I believe they will be found in the motion “beyond the center if mass” (displacement), because it is an outcropping of spin phenomena that may not be fully understood from an intuitive perspective. Questions, such as whether gyro motion “beyond the center if mass” (displacement) have acceleration (at any given time), Centripetal force, Centrifugal momentum, etc are not well explained yet.
Please review video number #9 entitled “gyroscope hanging from a long piece of string” at http://www.gyroscopes.org/1974lecture.asp, and consider that fast spinning gyros (with very slow precession velocity) revolve a distance away from the center of mass when hanging from a chord. On the other hand, slow spin gyros with very fast precession velocities do not revolve at a distant radius, but rather turn about the center of mass, in a manner similar to deadweight.
It does not appear that centrifugal momentum makes the gyro revolve away from the center of mass.
Harry, perhaps you can shed some light that explains this interesting phenomenon, displayed (somewhat shabbily) in the video. I have not been able to come up with an explanation that I can put to words at this point I time.
Best Regards,
Luis
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Harry K. - 12/03/2008 17:24:02
| | | Dear Luis,
Yes you are right, like many others here in the forum I try to help to expand common gyro wisdom and I also try to learn new things if there are something new to learn. Please appologize if you think I'm arrogant in some matters but this is not intended.
Nevertheless I have to state that your rocket ship example is wrong. It is not possible to move this rocket into a circular path by applying only one force at 90° to its constant velocity. To do that you need 2 counter acting parallel forces for rotating the rocket AND 1 continuous acting acceleration force. A spinning mass point is forced by design to follow a circular path and therefore not comparable to your space ship example. Or will your car follow a circular path if you drive straight on a street with constant speed and with constant acting side wind? And yes, the resultant velocity of your will increase because the reclined distance will increase under the influnce of side wind (assumed the width of the street is big enough for the diagonal path ;-) ).
Please reconsider again this issue.
Regarding video #9 I can only state that you see acting many different forces at the same time or alternating. This is a complex interaction of precession, nutation, tilting, potential and centrifugal forces and friction, i.e nothing out of the ordinary. ;-)
Best regards,
Harry
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Luis Gonzalez - 15/03/2008 20:16:08
| | | Dear Harry,
I should not be surprised that anyone would miss the true nature of the rocket example I presented in the context of this thread. When it comes to explanations of dynamic events, all it takes is the miscommunication or misinterpretation of just one word to confuse the meaning of the entire issue.
The only time there in nothing new to learn is when we stop learning. I suspect this can happen to everyone (including us). I am also convinced that simplification makes things clearer; so I will try to do that for our example.
Take an object moving through space at a constant velocity (can be a rocket or an arrow).
Attach a jet to one side of the object so that the jet always points to one side in relationship to the direction of the motion at any given time (i.e. despite the change in direction of the motion, the jet points at 90o to the direction of motion at that point in time).
You see, as the object is turned by the force of the jet, the direction of the jet also changes so that it still forces the new direction to change further (still at 90o, nothing like the side-wind on a moving car). Think about this a little.
Your car analogy is a completely wrong example, as a comparison (the word “wrong” can be so harsh).
First of all the car is restricted by friction in some directions but not in other directions because it is on wheels.
Second, and most important, the wind always blows from ONE direction; so if the direction of the car changes, then the force of the wind is no-longer affecting the car at 90o and ends-up pushing from slightly behind, more and more, as the interaction continues.
Finally, the car has a driver maintaining the course.
So, you see, the car does not really compare well to my example.
I think the language barrier continues to interfere with the clarity of our communication in this level of complexities.
The comparison of A) the object in space to B) a mass point (e.g. object spun at the end of a string) is a good comparison because both are being affected by a constant 90o force; one is “Jet Force” and the other is “Centripetal Force”. So you see I must insist on my original point. Please think it through well so we can get the facts correctly from here on.
Sometimes it is more difficult to see and think clearly under the public eye because we often feel a need to answer quickly, before we have fully understood other points of view. I often find it best to read over others points of view, until I understand them. In fleeting conversations it seems safe to give fast answers and appear correct at the moment (even when we are not). However the written word remains as record of what we state, long after we are through with the heat of a discussion, allowing people to always come back and se our errors. It is best to think well about the questions and answers we commit to in public record. It’s best to seek clarification before giving the “wrong” answer (such a harsh word).
Regarding video #9, my question addressed the difference in behavior of a fast spinning gyro Vs a slow spinning gyro. Both situations are similar but the behaviors are clearly different. Perhaps I just did not understand how you addressed that in your answer.
The main point is that the “Reorientation” of spin-deflection (precession) does come to an immediate stop because it does not have momentum (it is only a change to an existing direction). The only momentum in spin-deflection (precession) exists only in the motion “beyond center of spin mass” (“Displacement”).
Best Regards,
Luis
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Harry K. - 16/03/2008 12:35:38
| | | Dear Luis,
You see I do not accept you advice and reply quickly. :-))
It's curious, whenever you disagree with something I have stated, you mention "language barriers". I know that my English is really not the best, however, I never had understanding problems caused by "language barriers" with other guys here or in other boards. Nobody complained about language barriers but you did and do it in some situations.
Never mind I understand exactly your rocket example and I have to state again and again that it is wrong or false. Sorry Luis, it's not my itention to be harsh in any way but in Physics or Mathematics world there is only "right", "wrong", "false", "true" without flowery phrases.
Your main misleading assumption (I hope this is not too harsh?) is that the rocket will rotate around its center of mass under the influence of 90° side acting force and the constant velocity of the rocket. But that's not true! The rocket will certainly change its origin direction according to the resultant vector given from origin velocity and 90° side acting force. The rocket, however, will not rotate around its center of mass by itself into the new direction!
To achieve a circular path of the rocket, the side force must be rotated according every new resultant direction. Therefore you have to apply an additional torque to rotate the rocket in the desired new direction.
Exactly this rotation torque will be applied if a mass is forced by design to rotate around a fixed pivot. If there is no mechanically linked pivot to force a mass to move around a circular path, this rotation work must be applied in addition. I hope it is more clearer now!?
I have found a link with a nice example which shows what happens if two velocities acting at a body. It's a boat on a river. You can adjust the boat and water velocity and the direction of the boat speed. Change the direction and velocities with the mouse and move the boat with the mouse to the very left side and click "go".
Here is the link:
http://mysite.verizon.net/vzeoacw1/velocity_composition.html
Best regards,
Harry
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Luis Gonzalez - 17/03/2008 00:45:48
| | | Dear Harry,
Don’t feel offended, English is not my first language either. I have preferred to think that our disagreements are caused by misunderstandings (we both know that misunderstandings can occur even within the same native language).
I am not looking for flowery phrases, only to have my words understood but I believe I have not made my self clear enough. Maybe the following will do the job:
You said “To achieve a circular path of the rocket, the side force must be rotated according every new resultant direction”.
That is exactly what I have been trying to say that my example does. Remember that in the example the rocket/arrow has attached a jet that always redirects it sideways in relationship to its existing motion. An additional torque is not necessary because the jet turns as the rocket/arrow changes direction. The boat example is still NOT appropriate because the direction of the force of the water never changes; it is constant in one direction. The force of the jet does change direction.
I think you may be saying that it is not possible for the source of force (the jet) to change direction as the direction of the object in space changes. Is this what you are saying?
Best Regards,
Luis
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Luis Gonzalez - 17/03/2008 00:58:40
| | | Dear Harry,
There is something else I did not make clear.
If the jet is applied at 90o to the front end of the rocket/arrow then it will cause the necessary rotation as well as change to the existing direction. Applying the force to the center of mass will not cause rotation, but if the force is applied to a point ahead of the center of mass (and at 90o to the motion), then it should have the effect that I failed to explain well.
Best Regards,
Luis
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| Answer: |
Harry K. - 17/03/2008 10:13:08
| | | Dear Luis,
So we are finally in agreement that an additional torque is necessary to rotate the rocket into the desired new direction of movement caused by the side force? ;-)
This additional torque can be created certainly by applying a force beyond the center of mass, but the caused rotation depends on mass inertia, the distance to the center of mass and certainly the force size value. This caused rotation may not match with the required rotation speed and thus the only use of this torque would not be a good choice.
Anyway we should now stop to discuss these details further. I would be interested to hear your opinion to EDHs invention. Please visit the thread "EDH's gyro propulsion invention".
Best regards,
Harry
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Luis Gonzalez - 17/03/2008 16:42:30
| | | Dear Harry,
I am glad that I was finally able to express myself in a way that can be understood.
I have some thoughts about EDH’s invention but am focusing on what appear to be inconsistencies. I am pleased to see that many of the questions presented are appropriate to pursue the apparent inconsistencies (and more).
Hopefully EDH has brought the search for gyro-propulsion to a conclusion; however, I don’t think he has done it using the spherical spin-objects (even though the spheroids provide fascinating force interactions in 3 dimensions).
I still think that the subject I am pursuing will prove of theoretical value, long after the excitement over EDH’s machine is over.
Best Regards,
Luis
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