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27 November 2024 10:19
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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: |
Gyro-propulsion designers need to understand and utilize “J” the rate of change in acceleration. |
| Question: |
The rate of change in acceleration is the third derivative of the equation for distance and is scientifically referred to by the term “Jerk” and the symbol “J.” Newton did not include “J” in the 3 laws of motion, but he lay down the foundation for others to explore it. A=2S / ([t][t]) and J=6S / ([t][t][t]) where A=rate of acceleration, J=rate of jerk, S=distance, and t=time. Does this make sense?
The “rate of change of acceleration” (J) will provide the most useful measurements to explain and develop successful gyro-propulsion. People working on gyro-propulsion projects need to perceive and recognize the causes and effects of “J” in their design. They must also have the ingenuity to figure out how to make use of “J” effectively.
Where and when does “J” occur in gyroscopes? An object spinning at a steady rate is endowed with acceleration (centripetal “A”) by virtue of its spin. If the spin is accelerated, then the centripetal action shifts from “A” and becomes “J” but only for the period of time that the spin is being accelerated; there is a limit to how fast a flywheel can be made to spin.
How does “J” interact with gyro-propulsion? (Here is where Newton’s work pays off) Consider the following:
If 1) applying acceleration to a stationary object endows the object with velocity (V) at an increasing rate, then 2) applying “J” to a stationary object endows the object with acceleration (A) at an increasing rate (right?)
Therefore, if 3) applying acceleration (A) or torque (T) to the axis of a spinning flywheel endows the object with a constant velocity (V) at 90 degrees displaying no momentum (referred to as precession), then 4) applying “J” to the axis of a spinning flywheel endows the object with a constant acceleration (A) at 90 degrees & displaying no precession of its own (right??) (I have not yet found an official scientific term for this type acceleration).
Item 4 should explains the temporary lift (acceleration) produced by Sandy’s accelerated-system devices (in the upward part of the cycle), but it also explains why it is so short-lived (what Sandy refers to as “saturation”).
In a more complex interaction, “J” also explains why gyros display what appears to be a loss of centrifugal-action (centrifugal is not a force) when they get to a certain position. (I will let others try to explain this last item because, though interesting, it does not currently appear to have a direct effect on the type of design I seek.)
Limited propulsion can be achieved by well timed acceleration/deceleration of the torque applied upon the axis of the gyroscope flywheels. (The flywheels themselves may also be accelerated and decelerated strategically to enhance the effect.) However, this type of design is subject to brief windows of opportunity, in which to harvest linear acceleration, sandwiched in cycles that require re-adjusting the rate of spin of the flywheel (for the next opportunity to yield acceleration via “J”). This restriction produces long periods of energy costly non-acceleration (and short periods of profitable acceleration). Also, the magnitude of the acceleration produced is directly proportional to the magnitude of “J” requiring strong and heavy structures and actuators that yield relatively small propulsion.
Alternate designs that propose squeezing propulsion from the downward portion of the cycle also encounter 2 major obstacles. 1) Preventing precession eddies from steeling the full reaction to the downward thrust. 2) Mitigating the counter-acting forces when the flywheel reaches the bottom of the cycle. The solution to both of these obstacles is also found in the full perception and appropriate use of “J.” This (downward-cycle-thrust) strategy appears to have a proportional duration between the productive part of the cycle and the non productive portion of the cycle, and may have a lower index of wasted energy.
Whether one of these design strategies is better than the other or if a combination of both is possible, are theory and engineering questions. Some build high cost machines and others use mind experiments that lack concrete results for proof. It appears to me that a balanced strategy is necessary. Success will require good theory, knowledge of advanced materials, mechanical know-how, an electronic components engineer, a manager, legal capability, access to advanced actuators, some money, an appropriately equipped and supplied shop, and a great deal of ingenuity.
How can such a team be formed?
Spending the time and resources to build a machine that fails can take the wind out of the sails of effort and more if it self- destructs. The quandary whether to place effort on modified designs or to rebuild a stronger version of the last one can zap initiative and creativity, without which the effort grinds to a halt. However, learning from failures (our own and others’) is not limited to discarding what others have tried and failed at. As an illustration, many previous failures using propellers to invent the airplane did not deter two brothers from doing something slightly different and succeeding. Understanding the theory can allow us to see subtle things that can make the difference between success and failure.
I agree with Momentus that solving the problem at the bottom of the cycle is important but it is just as important to find the correct way to eliminate the precession-eddies. Thank you for your input Momentus, Nitro, Eric, and Sandy. Each one of your comments helps in one way or another. I also think that we can gain more by sticking to the subject without recriminations, or over-repetition of what has been stated before (it is as easy to make accurate reference to previously presented explanations).
I hope the intuitive theory is helpful. Tank you, Luis.
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| Date: |
27 August 2005
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Answers (Ordered by Date)
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| Answer: |
Sandy Kidd - 30/08/2005 07:49:16
| | | Luis & All
In an accelerated gyroscopic system angular momentum and centrifugal force (if you don’t’ like) diminish as soon as the gyroscope is rotated
It does not depend on position or anything else.
Gyroscopic torque is directly proportional to the rotation speed of the gyroscope, and as you all know, it is the leverage of the gyroscope around its own fulcrum that creates this effect.
Enough gyroscopic rotation (torque) and we have what I call “saturation”
All angular momentum and centrifugal force have gone.
This applies to each and every accelerated system.
Luis, I really do not understand what the benefits of this “J” factor are, all I know is that you are not going to get something for nothing.
I do of course realise that you will all go your own ways anyway, and I really do not see why you should listen to me, but remember what I have said, no matter what you do to a gyroscopic system, bend it, twist it, speed it up, slow it down, jerk it, kick it, or any permutation thereof you will not achieve inertial thrust.(except maybe as I originally did 20 years ago by sheer luck),
Then it took me years to figure out what was going on, basically because of my preconceived notions of what it was all about.
The answers were not what, or where I expected them to be.
The gyroscopic system is a prerequisite for inertial drive, but it is just not capable of doing it all on its own
Regards,
Sandy.
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| Answer: |
Gordon Vigurs - 12/09/2005 10:27:54
| | | Not so much an answer, more a comment:
I attended a lecture by Bernard Laithewaite at Farnborough 6th Form College in the early 1980s, and I must admit, he was an excellent showman. I remain sceptical, but there is an unsatisfactory feature in the equations of motion of a rigid body, which is never referenced in the text books, and has always troubled me.
The principle of angular momentum is an extrapolation of the principle of linear momentum, and using it, the classical gyro equations are easily derived. On closer examination, all we are doing is summing the effects of the linear momenta of all the particles of the body. Thus the principle of angular momentum actually deals with translational momenta of particles constrained to travel in curved paths by the forces holding the body together. Nowhere in the derivation is there any requirement for the particles themselves to rotate,
This is clearly an inconsistency, as compatibility of the structure must force the particles to rotate, and Newton’s Laws of motion has nothing to say about the effect, if any, of spin. There is potentially a theoretical basis for spinning bodies to exhibit properties not accounted for by linear momentum alone.
Now if we consider the same problem applied to fluid flow, we are similarly dealing with particles of fluid, which distort in the flow, the principle of linear momentum is again extrapolated to a large mass of particles, but this time with a strict requirement for the particles to fit together (the equation of continuity) so that voids cannot appear in the fluid. The continuity requirement makes a tremendous difference to whether the fluid particles themselves rotate.
The classical example is the vortex. At the centre, fluid particles rotate as a rigid body, the velocity distribution is parabolic and the flow field is known as a forced vortex. Outside the core (the eye of the storm) the fluid particles retain their orientation and the velocity varies inversely with distance from the centre; this is known as a free vortex.
This is a kinematic model. In practice the core flow is caused by viscosity, such that any closed path enclosing the core will be associated with a net friction torque whilst paths which do not enclose the core have no net friction torques; the only fluid force acting is normal pressure.
The important point is that different solutions can be obtained when continuity of the medium is taken into account. The particles of a rigid body cannot possibly maintain their orientation with respect to inertial space as the body rotates. Either the spin of individual particles has no effect on the motion, or there is a possibility we are missing something.
Nowadays work on propulsion is conducted entirely by experiment, and largely derided by theoreticians who are generally unaware of the limitations of the theory.
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| Answer: |
dave brown - 13/09/2005 04:08:56
| | | does this help with loss of centrifuge?
Loss of Flywheel spin due to precession.
As the shortest/fastest path between 2 points is a straight line,
the change of orbit is a longer/slower path.
due to the action-reaction at the top and bottom of the Flywheel-Shaft points, the orbit of the matter in the Flywheel is changed on the first half spin.
--- that puts a torque on the Shaft about the horizontal, precession.
then there is the blah blah of it.
Gyro Nutshell
-0--.
1) the Mass of the Flywheel will let us calculate the weight force.
2) the Shaft it spins about, and the Pedestal, will allow interaction with the Normal force.
--- a non-spinning Flywheel will torque the Shaft down at the point connected to the Pedestal.
---- 5lbs @ 2 ft gives 10 ft/lbs of ccw torque.
3) a spinning Flywheel will change the torque from ccw of the Shaft about Pedestal, to cw of the Shaft about dead-Center of Mass of the Flywheel, to keep it's axis of rotation unchanged.
--- the weight of the system is felt straight down on the Pedestal point, as it is no longer the center of torque, but tangential to the torque. the Normal force is clearly seen doing it's thing, opposing gravity. (the Normal force may not equal the system's dead weight force, as there is a downward vector of system movement.)
--- the inertia of the Shaft vs. the Inertia of the Flywheel dictates how much the axis of rotation is deflect ccw in direction of the applied force.
--- the Normal force is 5lbs at 2 feet = 10 lbs. the Flywheel is experiencing @10 ft/lbs of torque.
a spinning Flywheel wants to fall straight down. the Shaft on the Pedestal is what allows us to experience this spectacular view of the laws of forces in action.
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| Answer: |
Luis Gonzalez - 16/09/2005 02:38:01
| | | Gordon,
From a theoretical point of view, it is interesting (maybe even important) to ask whether spin introduces factor(s) that should be classified as other than new dimension(s) in respect to time.
If the gyro phenomena is something completely new and different, then someone needs to start building a consistent theory that can explain all (or much) of the phenomena observed in experiments.
However, if spin actually introduces only a new dimension in respect to time, then the existing realm of mathematics is sufficient to support the resulting theory within “n” dimensions. In such case, an acceptable strategy is to include the additional dimension(s) in a way that does not fly in the face of the remaining body of observed and proven science, and then to evolve the math, physics, and mechanics in a rational way.
What I am saying is that it may not be necessary to introduce a new set of unknowns to satisfy observed phenomena. The complete explanation of the phenomena that contains the least number of new variables and explanations is likely to be the most elegant and probably the best explanation.
I agree that physics books do not cover well certain areas that are not often or deeply used in industry. Gyros and the rates of change of motion that go beyond acceleration are among such areas of knowledge. Current technology has limited use for these kinds of knowledge and therefore does not explore it.
You spoke of fluids-flow and it brought to mind the behavior of super-cooled liquids (referred by some as quantum liquids). When a normal liquid is stirred in a circle it forms a single funnel. When a quantum liquid is stirred it forms many smaller funnels. Not wishing to delve too deeply into this phenomenon, I will let others ponder the thought. However, I believe the mathematics behind this quantum behavior may have a common thread with the behavior of gyros.
One such promising connection may be found in a mathematical paper that proposes a set of equations that bridge the apparent gap between Newton and Quantum mechanics. I don’t have the source handy at the moment but would be happy to share it with interested parties. If the mathematics paper is correct and the behavior of quantum liquids is as stated, then the application of Newton to gyro behavior may be much less farfetched than the path I have just hinted at.
Regard,
Luis
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