Home : Gallery : History : Uses : Behaviour : Maths : Forum : Propulsion : Links : Glossary
Main Forum Page

The Gyroscope Forum

20 May 2024 14:35

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.

Search the forum:  
 

Question

Asked by: Luis Gonzalez
Subject: Building Successful Propulsion Devices
Question: Designers must be able to estimate the expected thrust (at minimum) of proposed propulsion-devices. How can one calculate the thrust of a gyro-propulsion design and why is it important to estimate?
First, trial-and-error innovation requires enormous funding and resources. It is possible to obtain initial results without well-thought-out planning. However final success requires a tradeoff between rigorous planning and an extraordinary number of trials and failures. (Tomas Edison required over 5,000 experiments to invent the incandescent light-bulb. How many gyro-propulsion experiments have been completed thus far, and at what cost?) Accidental discovery is common but must be followed by smarter resources to achieve desired goals. (Finding that bread-mold destroys bacteria was accidental, the creation of antibiotics was not.)

How does one estimate the expected thrust of a device?
My device is a 2 cycle design. The first cycle uses precession to position the flywheel at a higher point. The second cycle produces thrust by returning the flywheel to the original position (it is key not to allow downward precession during this part of the cycle), thus creating a counter action (equal and opposite reaction) on the part of the device as a whole.
My thrust-estimate is based entirely on the downward motion of the flywheel. The distance from the top-most position of the flywheel to its bottom-most position is the length of the trust. (Yes, I must mitigate or prevent the counter-action that can be caused as the flywheel is brought to a stop at the bottom of the cycle and thus cancel the benefit of the downward motion. That is where the cleverness of the design comes into play.)

The distance of the thrust is determined by the length of the axis that attaches to the flywheel and the angle of rotation during the downward thrust. The Force = F of the thrust is determined by the mass = M, times the acceleration = A applied during the completion of the downward flywheel motion through a distance = S. The distance (S) from the top position to the bottom position is equal to S= (1/2) (A) (t) (t) where A=acceleration rate, and t=time. From this common equation we can derive the rate of acceleration during a specific stroke, it is A = (2S) / (t) (t).
Therefore if I can determine the displacement and duration of the downward stroke (based on the dimensions of the design and the torque of the motor, spring, or other actuator, etc) then I can calculate the acceleration that the flywheel will be subjected to, during the downward motion.

To obtain the ongoing rate of acceleration (as opposed to acceleration during a single stroke) I must estimate the duration of thrust in relationship to the full cycle. This ratio is then multiplied by the acceleration of the downward-stroke. In my design the acceleration lasts half the time of the full cycle so I must multiply the calculated acceleration by 0.5. The result can be adjusted for vibration (caused by dead weight of the flywheel-frame and such). The final acceleration is then multiplied by the mass of the flywheel providing a close estimate of what’s expected to be the net Force (F=MA) produced by my design once it is built and operated.

To determine how much weight or lift my machine may deliver I need to estimate the weight of the entire apparatus (including flywheels) and multiply that by the acceleration of gravity (32ft/sec.sec) yielding the gravitational force that keeps my device solidly on the ground. The ratio of, the force produced by my device, to the force of gravity on my device, should provide a percent-of-lost-weight (or the rate of upward acceleration).

The next question is; how can I improve the upward force of my device? I can:
a) Increase the ratio of the mass of the flywheels to the total weight of the device
b) Increase the distance of my downward thrust
c) Reduce the time (t) that it takes for each stroke/cycle to complete
d) Or reduce the overall weight of the device.

The strength of the components and the power of my motors etc limit how much I can increase the mass/weight of the flywheels.
The distance of the downward thrust is limited by the overall size (not weight) of my device.
However, I can reduce the overall weight of the device by innovative use of modern materials and engineering. With the right materials and technology I can also reduce the time of each stroke and this provides the greatest gain because, as I reduce the duration of each cycle (t), the acceleration increases exponentially (while all the other factors provided only linear benefits).

Q. How would you estimate the propulsion thrust of your design?
(If you can’t come up with a good estimate, it may be time to re-evaluate your design.)
Thank you,
Luis
Date: 22 August 2005
report abuse


Answers (Ordered by Date)


Answer: Luis Gonzalez - 22/08/2005 03:57:07
 Glen,
I want to add this question to the "Gyroscopic Propulsion" area (not this one), but I cn't seem to get it to go there.
Thanks, Luis


Report Abuse
Answer: Nitro MacMad - 22/08/2005 18:19:27
 Dear Luis Gonzalez,

Bloody good letter, but be careful. If you want to claim anything as originating from you in a patent in the future you won’t be able to if it is disclosed on a public forum like this.

The reason that no one seems, thus far, to have cracked “impulse drive” is contained in the last six words of your fourth paragraph.

Once it is made you will find that your calculation of net force is about five to ten times what you will actually achieve. This is due to the “dead force” period that occurs while the process you call the “key”, mentioned in brackets, in your second paragraph occurs. That, “key”, process seems so simple but, be warned, is a right basket (kind of devise, held on your right hip, for keeping fish in – I think).

Keep it up but GET MAKING.

Kind regards
NM






Report Abuse
Answer: Sandy Kidd - 23/08/2005 07:47:54
 Dear Luis,
Notice from your posting that you are just as keen as ever, and full of confidence for your device. I know this feeling well.
I cannot help being a kill joy but I think it would be much better, more sensible and much more important to get something to work first. i.e a device which produces some thrust (any level of thrust) rather than bother about its output at this early stage
That can come later.
I have stood where you are now standing, hundreds of times over the years and have experienced complete disillusionment and disappointment on most of those occasions.
The depression wears off after a while, and if you are a bit stupid like me, you mount the roundabout once more, just to get more of the same.
You will discover at the end of the day that any thrust produced from a gyroscopic device comes from sources not normally taken into consideration.
Mother Nature is very simple in almost every thing she does, but she can be a cunning bitch when she comes to hide the answers..
I shall be genuinely pleased for you, but more than a little bit surprised if the set up described delivers anything at all, except, as in the case of a few of my machines, maybe by default.
You will find out to your dismay that no matter what you do to manipulate a gyroscopic system ON ITS OWN, it will not produce inertial thrust.
Best of luck
Sandy.


Report Abuse
Answer: Momentus - 24/08/2005 11:36:35
 Hi Luis

“Up Like a Gyroscope, Down Like a Dead Weight”. The first device I built, which started my association with Prof Laithwaite worked on just such a principle. He was impressed with the principal and asked to see the shattered remains!!!

Referring to my previous posting
http://www.gyroscopes.org/forum/questions.asp?id=350

The walking gyroscope embodies this principle. It precesses forwards in a series of arcs, and then is pulled back in a straight line. This moves the centre of mass, but does not accelerate the system as a whole.

The key feature is as you say, to stop the moving dead weight, retaining the acceleration. In my first device, this involved a very ingenious method of switching the gyroscope on and off, much mechanical complexity, and one test run to break it beyond repair. Did it give thrust? Don’t know too busy dodging flying parts. But I was positive enough that I saw the needle shift!!!

I have experimented with the walker, stopping the returning “dead weight” at one support point, causing it to arc in precession. That seemed to absorb the force, but the whole thing started to get too complicated. The walker was built to demonstrate displacement, not thrust.

If you have solved the “stop the dead weight” problem, then you will get thrust, not just displacement. As Nitro says JFD.

Hope this helps, good luck.


Report Abuse
Answer: Eric James ----- - 25/08/2005 04:16:20
 Luis,

There's lots of information on calculating thrust coefficients available on the web and in libraries and book stores. Mostly, the available material is in relation to continuos thrust systems (like aerospace applications). You would need to calculate an average thrust pressure for your concept in order to use the formulas.

Basically, if you know how much thrust you get with each thrust cycle (by experiment and measurement, preferably) and know how long the thrust to setup ratio is, you can calculate your average thrust.

Then just plug in the numbers for your mass and such and viola! You have your answers.

Don't forget to include negative thrust in your average though (like your "counter-action," where the gyro stops at the bottom).

Eric

Report Abuse
Answer: Luis Gonzalez - 27/08/2005 22:10:24
 Te 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.

Report Abuse
Answer: Eric James ----- - 28/08/2005 04:44:50
 Luis,

Momentum equals mass times velocity!
Force equals mass times acceleration!
Yank equals mass times jerk!
Tug equals mass times snap!
Snatch equals mass times crackle!
Shake equals mass times pop!!
Credit: http://math.ucr.edu/home/baez/physics/General/jerk.html

Obviously "jerk" is important to the comfort of any passengers and delicate equipment, but exhibiting jerk doesn't mean that you have a net acceleration (jerk can be reciprical). This is actually an important consideration in high performance internal combustion engines (ICEs) as it affects the wear and tear on the pistons, wrist pins, connceting rod bearings, crank and whatnot.

Jerk is commonly incorporated into a variety of inertial propulsion concepts too. In fact, it is THE principle upon which many have mistakenly relied upon. Usually, it is erroneously considered that a soft jerk on one end of a cycle provides less acceleration than a hard jerk on the other end. Therfore, it is incorrectly reasoned to exhibit asymmetrical acceleration properties (accelerates more to the left than to the right, or whatever).

The reason why it doesn't work is because a soft jerk requires more time than a hard jerk. Mathematically, they are the same in terms of total acceleration.

Anyway, I think that you are too quick to gloss over a major obstacle to your concept in the hope of "improving it" away.

I'm sure you're aware that the "problem at the bottom of the cycle" is simply an equal and opposite reaction to the thrust at the top of the cycle. It's a reaction delayed by distance, but a reaction nonetheless.

There's an interesting property to such delayed reactions. The boundaries of the container might move, but the center of mass for the system stays put. The center of mass adjusts rearward along with the mass that you accelerated against moves rearward. There is no actual net movement of the entire system's center of mass.

I don't think you can "improve" this reaction away. It's a fundamental consequence to your action at the top of the cycle.

I'm not quite certain what you mean by "precession eddies." Are you referring to direction changes of precession due to torque angle changes? Perhaps you are referring to the tendancy for the the pivot to be spun around by the gyro? Or is it something to do with nutation perhaps?

Eric

Report Abuse
Answer: Luis Gonzalez - 28/08/2005 05:47:25
 Eric,
I am using term “jerk” strictly to represent the mathematical first derivative of acceleration, second derivative of velocity, and third derivative of distance. In this context the existence of “J” is a necessary prerequisite to change from one acceleration vector to another acceleration vector (including zero).
You can tell from what I wrote that I am NOT proposing to use different levels of “J” (soft & hard) to create propulsion.

I am willing to share my view of the basic mathematical theory of the observed phenomena. However I am not willing to share my design solutions to the major obstacles (that you perceive as glossing over).

“Precession eddies” are another way of calling what Nitro calls Nitro’s first law.

Thank you,
Luis


Report Abuse
Answer: Eric James ----- - 28/08/2005 08:44:19
 Luis,

Okay. I was just concerned by your apparent nonchalant attitude to the reaction force at the bottom. I feared it might indicate that you didn't understand the consequences. I'm happy to read that I was in error.

Still, I don't see how jerk is all that important. It seems to me that it's not the rate of acceleration change that's important, but rather it is that a permanent (overall) acceleration change is achieved.

Anyway, I like the term "precession eddies." It's a good one. I suppose that you are referrring to chaotic forces causing unexpected precession in unexpected or undesirable directions (derived from Nitro's first law). Is that fairly accurate?

I would be very interested in seeing your mathamatical analysis of your theory (if you're willing to share it). Include lots of explanatory text though, as my math skills are a bit rusty (okay, actually they are rotted to the Earth). I'm a quick study though, as long as the logic flows.

Eric

Report Abuse
Answer: Nitro MacMad - 28/08/2005 09:56:29
 Dear Eric, Luis and other rotor heads,

I suppose because I started from a pendulum I never made a machine that optimistically pointed towards the stars with the hoped for direction of thrust upwards. My Heath Robinson contraptions have all had their main shaft horizontal. It’s a lot easier to see and video the various reactions – and in the name of big hairy caterpillars there are enough of those to stop you being bored for a while- with a trolley mounted horizontal machine.

The trick is to get as much as possible of the first part of the cycle free of opposite while it is accelerating - with or without “J”, to choice – and then use the deceleration AND the following opposing acceleration on the second, return stroke as your drive.

When you catch up please give me credit (What do you mean – “big head”? I’ll have you know, that if I grease it enough I can still get it through a standard door! Stings the ears a bit though!).

Kind regards to all
NM

PS You guys in America must look up W Heath Robinson on the web. My all time favourite is the little old lady saved from drowning in the Sound of Mull.

Report Abuse
Answer: Luis Gonzalez - 28/08/2005 21:16:22
 Eric,
Take a 2 stroke cycle scenario. When the flywheel is forced to “precess” upward, it has an angular velocity through an arc, which produces little or no opposite reaction (because the reaction takes place against the accelerating actuators (at 90 degrees) that fomented the precession, etc). There is no lift (propulsion) to be had from this activity (else our work would be done). However, if you do more than just force “precession” (by simply applying acceleration) and instead apply increasing acceleration “J” then you will not only get a simple upward precession (that has simple (V) velocity), but you will get a hyper-precession that has upward acceleration (A). Is this clear? Does it ring out loud? (Assuming the theory has produces positive empirical results; has it?) If upward acceleration is indeed achieved, then we have a component of propulsion (F=MxA). Do you now see why taking “J” under consideration may prove helpful during the upward cycle?
The term “precession eddies” is derived form Nitro’s first law, but is meant to describe something perceived as less chaotic and unexpected than it has been thus far.

The basic mathematics for the upward part of the cycle is not as complex as one may think though it requires modest innovation (without getting too wild). Simply divide “J” by the angular momentum of the flywheel and you have your upward acceleration for that upward half of the cycle. The acceleration (A) of the resulting hyper-precession = jerk divided by angular momentum of the flywheel. A=J/L. It doesn’t last very long, so you have to figure out the duration of “J” over the distance of the arc etc.
The math for the downward stroke of the cycle has already been provided in this thread. However, it does need to be extended to include what happens at the bottom of the cycle, and the how it handles the precession eddies……. (Need I say any more?)
Buy a couple of books on the basic college Physics. Spend some time on the basics of motion (and how mathematical derivatives and integrals apply). Focus on the formulas for motion and for calculus derivatives and integrals. They are not very long and not hard to get used to.
Thank you, Luis


Report Abuse
Answer: Eric James ----- - 29/08/2005 04:22:04
 Luis,

Okay, good. Now I see why J is important to your concept.

It looks to me though that your whole concept hinges on the affirmation of the proposal that a gyro ALWAYS precesses around the point of applied torque, even in isolated systems. Is this correct?

What I mean by this is that if this is not true, in an isolated system your "upward" precession will have an equal and opposite reaction as it pushes against the frame that it is precessing against, meaning that there is no net movement of the center of mass.

Eric

Report Abuse
Answer: Luis Gonzalez - 01/09/2005 03:14:14
 Eric,

My concept?
Using the downward part of the cycle to produce gyro-propulsion has been around since the 1970’s or before (I think it was popularizes by Professor Eric Laithewaite and others).
Using the upward part of the cycle probably came soon after or simultaneously (I think that was popularized by Sandy Kid and others, but I am not sure).
These 2 items appear to be basic “principles” of gyro-propulsion. I currently know of no other way to create propulsion (though there are a number of variations of these 2 principles, including Momentus’ walker). If a design is not based on either of these 2 principles than it is introducing a new (even more innovative) solution to gyro-propulsion (that I have not heard about); on the other hand it may require analysis to determine why it should (and testing to prove whether it will) produce propulsion.

My “concepts” are embodied in designs to implement the principles (that I know about), and to overcome the challenges encountered so far (i.e. how to build a self-contained device with the necessary attributes and dynamics).

As of yet, I have found no literature on a gyro-propulsion-model that is based on a foundation of science discipline (physics, math etc). For that reason (plus the unresolved issue whether gyro-propulsion defies Newtonian principles), I have written an analysis on how “J” links the equations of motion to propulsion (during the upward part of the cycle). The starting point is that centripetal acceleration (A) “changes” when the flywheel’s velocity (V) “changes” (rate of RPM), and that both “changes” are directly proportional to each other. So “J” (change of A) was there all along but nobody appears to have ever mentioned it.

A solid theoretical foundation needs to be based on observation of phenomena and how it fits within known science. Good theoretical foundation brings credibility to the effort and is a tool in helping to evaluate designs. We should not have to build gyro-propulsion devices by luck and chance; rather we should build them by design based on scientific principles. This and all other well thought-out and explained principles are helpful to the effort.

Regarding your second sentence, I am not sure what you mean by “isolated system.” If you mean absent of gravity (as in space) then the answer is a qualified (assuming I understand you correctly) because gyro precession appears to occur around an axis line more so than around a point).

Finally, precession does not have an “equal” and opposite reaction (at 180 degrees). The action that creates precession does have an equal and opposite reaction but both the action and the reaction occur at 90 degrees to the precession.

Thank you, Luis

Report Abuse
Answer: Sandy Kidd - 01/09/2005 10:36:43
 Dear Luis
About 25 years ago, utilisation of up and down part cycles of gyroscopic action seemed at the time to be logical and obvious paths to tread
As you state that was some considerable time ago
I personally advocated the “in and out” (with a bit of up and down) rather than the more “up and down” bias.
I am sorry this sounds like waffle.
Anyway not to put too fine a point on it, it’s a waste of time, and there are no gains to be had in any of these directions.

Nearly 20 years ago I built a rather expensive device which would definitely bring the whole deal to a successful conclusion.
This wondrous device was built with 4 spherical gyroscopes.
The spheres were each constructed in 2 parts (what else - hemispheres)
The hemispheres were supported on a shaft which was fitted with needle bearings and a Spraag clutch
The rotation axis of the shaft was designed to rotate parallel to the rotation axis of the machine. Without a sketch this is not so easy.
Let me just say that for one half a revolution the sphere was a gyroscope, and the other half a dead weight.
The spheres worked perfectly, oscillating up and down beautifully, or as somebody said “up like a gyroscope down like a brick”
Trouble was, it took 5 weeks of severe modification to get a few ounces out of the machine.
I deliberately altered the device to operate in a way that I knew would provide some thrust.
This test was accompanied by another certified successful lab test.
These modifications virtually eliminated any attributes offered by the split spheres.
Complete waste of time and money. £10,500s worth.
Thank you, Sir Isaac.
A point of interest. Later, in London, when that particular machine was the subject of discussion with Eric Laithwaite, I was accused of withholding information from him. Our association ended there.
However, that is a story on its own, and not for here.

Anyway none of these “ups and downs”, or “ins and outs” in any way shape or form will produce measurable thrust
You may, by default get something, but it is not going to take you to Mars.
However any gains at all, at this game, are a bonus.
I started with a defaulting, but consistent, device 20 years ago.
With a bit of laser technology I could have had all the answers right there and then, but thanks to the university’s attitude towards “this disreputable device”, I have had to “knife & fork” the problem, bit by bit, over the years.
In any event the answers were in a totally different and unexpected place.
The task has been similar to looking for a needle in a haystack, without any guarantee that there was ever a needle in there in the first place.
Nobody said it would be easy.
Sandy.


Report Abuse
Answer: Luis Gonzalez - 03/09/2005 14:54:12
 An effective gyro-propulsion device requires spinning of flywheels and taking advantage of phenomena such as precession (among others). This requires motion(s) determined by the design. What are those motions?

Deciphering what is meant by ‘ups-and-downs, or ins-and-outs’ that are a waste of time, or that ‘answers are in different and unexpected places’ is guesswork that leads no closer to an answer. Answers will NOT come from any of the inventors. We must each discover what works by following our own intuition and knowledge. Which activities and experiments to repeat, and which ones to avoid, is your own choice.

People working toward the same goal are either collaborating or competing. We can only imagine the reasons for postings in this website; to clarify questions, to seek collaboration, etc.

I hope that some of us may come to understand the principles of internal-propulsion by sharing questions about the underlying science. We are all of course aware that sharing design information requires collaboration agreements.
Thank you, Luis

Report Abuse
Answer: Sandy Kidd - 06/09/2005 15:37:14
 Luis,
I am an engineer, but I am called an inventor, a pretentious term perhaps, and a label I do not care for.
This term is usually applied to someone who has designed or developed something which is unique, novel, or clever.
It is a term also applied to someone who aspires to design or develop something which is unique, novel, or clever.
What have you got against inventors, Luis, we would have nothing without them?
History is full of them, both successful and otherwise, and if you discover how to do this thing you will like it or not, also become an inventor.

Non disclosure agreements and collaboration agreements are non workable and equally worthless.
Usually ends up with someone being severely ripped off, and that someone is invariably the major intellectual contributor.
Words, paper and signatures are cheap.
Loyalty and trust are something else, and very hard to find.

Some time ago I made a statement to NM on this forum that a patent was not worth the paper it was printed on, and at that time, was for very good reasons.
I apologise to NM in this instance, as it now appears, that for other very good reasons, it could be in my best interests to do so.
I am nearing the end of preparing my first patent application in 20 years.
Study my patent Luis.

Luis, you have classified my statements as guesswork.
You state that the answers will not come from inventors.
Did you mean the inventors will not give you the answers?
I am sure this is not what you mean, but if it is, why should they?
I have tried to convey over several months salient features which I would describe as fundamental issues. Probably all read without comment, but nevertheless all registered on the forum and all provable.
I felt it necessary to warn folks not to treat gyroscopic systems in isolation, unless they specifically want to spend 10 or 15 years chasing moonbeams.
At least consider what I have stated, no more, no less.
Any deeper than that, I am not prepared to go.
Your postings relating to the “J” factor, machine timing etc. etc are all guesswork.
You don’t get something for nothing.
You want to discuss the theoretical output of imaginary devices.
If the operating parameters are unknown, the output is totally irrelevant, unless of course your name is Pournelle and you wrote a book called “A Step Farther Out”
Regards,
Sandy.


Report Abuse
Answer: Sandy Kidd - 06/09/2005 15:37:20
 Luis,
I am an engineer, but I am called an inventor, a pretentious term perhaps, and a label I do not care for.
This term is usually applied to someone who has designed or developed something which is unique, novel, or clever.
It is a term also applied to someone who aspires to design or develop something which is unique, novel, or clever.
What have you got against inventors, Luis, we would have nothing without them?
History is full of them, both successful and otherwise, and if you discover how to do this thing you will like it or not, also become an inventor.

Non disclosure agreements and collaboration agreements are non workable and equally worthless.
Usually ends up with someone being severely ripped off, and that someone is invariably the major intellectual contributor.
Words, paper and signatures are cheap.
Loyalty and trust are something else, and very hard to find.

Some time ago I made a statement to NM on this forum that a patent was not worth the paper it was printed on, and at that time, was for very good reasons.
I apologise to NM in this instance, as it now appears, that for other very good reasons, it could be in my best interests to do so.
I am nearing the end of preparing my first patent application in 20 years.
Study my patent Luis.

Luis, you have classified my statements as guesswork.
You state that the answers will not come from inventors.
Did you mean the inventors will not give you the answers?
I am sure this is not what you mean, but if it is, why should they?
I have tried to convey over several months salient features which I would describe as fundamental issues. Probably all read without comment, but nevertheless all registered on the forum and all provable.
I felt it necessary to warn folks not to treat gyroscopic systems in isolation, unless they specifically want to spend 10 or 15 years chasing moonbeams.
At least consider what I have stated, no more, no less.
Any deeper than that, I am not prepared to go.
Your postings relating to the “J” factor, machine timing etc. etc are all guesswork.
You don’t get something for nothing.
You want to discuss the theoretical output of imaginary devices.
If the operating parameters are unknown, the output is totally irrelevant, unless of course your name is Pournelle and you wrote a book called “A Step Farther Out”
Regards,
Sandy.


Report Abuse
Answer: Luis Gonzalez - 07/09/2005 02:55:43
 Inventors are among the smartest contributors to civilization and they would not be smart to openly giveaway their design solutions (answers).
Collaboration should be agreed upon with trust, often to complement one’s skills, and in some cases to cause necessary changes in the course of matters that can not happen otherwise (especially if profit and credit are not as important).
I am happy to hear that you intend on filing a patent (it’s about time), and I will certainly examine it when it becomes available; do let us know.
I have no way of knowing whether or not your statements are guesswork any more than you can know whether mine are. What I do know is that it will take guesswork to figure out some of your statements.
Thank you for trying to save the rest of us time spent arriving at similar results as you have by following a similar path. We can’t exactly expect you to reveal all your secrets and some of us are bound to be both more and less smart than you Sandy.
I hope to see your new patent soon and hope we don’t need to waste our time writing notes such as this again. I would much rater spend time deriving mathematics for the concept as I see it evolve.
Thank you, Luis


Report Abuse
Answer: Sandy Kidd - 07/09/2005 08:07:32
 Dear Luis,
They say you can take a horse to water etc.
When my older brother, a much smarter than the average engineering design consultant, and in the company of a university professor, saw my first machine deliver its consistent 1 pound of inertial thrust, he made the comment that it could take me 10 years to develop that machine to a point of commercial viability.
I told him that he was talking rubbish.
He was wrong.
It has taken me 21 years.
Regards,
Sandy


Report Abuse
Answer: Luis Gonzalez - 09/09/2005 02:19:15
 Sandy,
There’s a lot of leading horses to and from water etc.
Now that you have arrived, how long will your new invention take to commercial viability?
Can you say how much thrust it produces?
Thanks,
Luis


Report Abuse
Add an Answer >>
Website. Copyright © 2024 Glenn Turner. All rights reserved. site info
Do not copy without prior permission. Click here for gyroscope products