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The Gyroscope Forum |
23 November 2024 18:23
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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: |
Albert Druid |
| Subject: |
the basics |
| Question: |
Been reading thru this interesting website. The extreme passion stopped - I see fewer postings. You all gave up or learnt what you came for.
Nitro Macmad asked a good question on posting id=306. And no one except one man tried to answer. Has anyone learnt enough to give it a clear answer? Nitro asked - why a gyro goes swiftly to a set speed when released - one might reasonably expect it to keep accelerating under the constant force of gravity? Hint - The best answer I seen starts by answering why precession is always orthogonal to gravity?
Good wishes.
Al Druid |
| Date: |
6 August 2010
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Answers (Ordered by Date)
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| Answer: |
Glenn Hawkins - 06/08/2010 04:26:13
| | | Hi Albert,
Precession speed is determined by the angular momentum in the flywheel, acted upon by gravity. Gravity is a constant and angular momentum is at any instant near constant. The two constant forces oppose one another so they must produce a constant precession speed that is the balance between each.
Was it Nitro, or Metcalf (sp?) who said precession begins as instantaneous as it can get? Well nothing can be instantaneous, but he seemed almost right. Precession could not begin faster than 32 feet per second, which is the speed of gravity. When you divide 32 feet by 1/20 of an inch or so, which can be the distance of the wheel’s free fall before it catches and converts to precession, and then equate that high number to the slower speed of eye fixations (only 25 per second) then precession will seem instantaneous to the poor deceived necked eye. Good of you to observe the question. Post more often, Albert.
Regards,
Glenn
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Albert Druid - 07/08/2010 17:10:28
| | | Good start Glenn,
Precession speed combines constant force of gravity with constant resisting momentum in the flywheel. Two constant vectors result in a fresh constant vector in a new direction - so the textbook explanation goes - precession = T / L - toque divided by momentum.
Gravity don't have constant speed of 32 feet per sec - it has constant acceleration of 32 feet - per sec - per sec - producing ever increasing speed. On the other hand the flywheel has constant speed only. They aren't truly opposites in the same league - torque acting on momentum.
Your explanation is like the one in the books to satisfy math students but is not fully clear - it don't explain details on the way torque acts on momentum or how exactly they oppose one another or why momentum, without acceleration, is considered as a force.
You took a good stab at it but it needs more deeper clarity. lets see if someone else can give more details.
Thanks for trying , and keep up the good work.
Al Druid
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Glenn Hawkins - 07/08/2010 21:25:02
| | | Albert, you say, “Gravity don't have constant speed of 32 feet per sec - it has constant acceleration of 32 feet - per sec - per sec - producing ever increasing speed.“
No, no. That is not the point. Increasing speed in a free fall increases momentum. Momentum is built upon itself inch by inch and collision force becomes greater and greater. . . but the force of gravity is constant. It doesn’t increase at all, except in varying elevations.
You say of my explanation, “. . . it don't explain details on the way torque acts on momentum or how exactly they oppose one another or why momentum, without acceleration, is considered as a force.”
This that you now ask is totally new. These questions weren’t ask in the thread. I can give the new explanations, but exact and clear explanations as I see them in my mind would requires a dozen pictures and many, many pages of writing. Even to understand the explanations as I see them would require several hours if not days of work and reasoning by the reader. These mechanics I speak of aren’t in the books. I invented them. And in the end, I think nobody really cares.
I gave the explanation to the question ask. If anybody gives a different explanation (I’m not talking about structure and languish, but mechanics) then the explanation will be wrong. You have what you ask for.
Thanking me for trying might be out of order. I didn’t try. I succeeded. I wish everything else in my life were this easy.
Be nice, Albert
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Momentus - 12/08/2010 15:23:57
| | | Hello Albert, welcome to the world of the shed dwellers.
My passion has not diminished, but has been overwhelmed by a screaming frustration.
Instantaneous is a specific word in physics with a specific meaning. It best translates into ‘simultaneous’ in the real world. A Gyroscope is precessing if a torque is present. The smallest movement has an accompanying torque, and of course visa versa. As the one cannot exist without the other, then they happen instantaneously.
I can give an example – imagine a circular billiard table with a ball bouncing back and forth across the centre of the table. As it moves over the centre spot it is deflected thro a small angle then goes on to hit the opposite cushion, bounces back and as it crosses the centre spot is given an equal and opposite deflection. Keep this up and the point of impact will move around the table. Call this movement around the table precession.
The angle of deflection is governed by two factors, how hard the ball is struck, and how fast it is travelling. This in turn will determine the speed at which this movement (precession) will occur.
If you are hitting the ball it is ‘precessing’ If you stop hitting the ball it is not. Turn the flat table into a hollow sphere with four equally spaced balls, deflect at the north and south poles and you will have a basic gyroscope.
Which will also answer Nitro’s question.
One final point. A gyroscope orbiting a tower is doing two things.
a) It is orthogonally rotating about the c of g of the spinning mass, an angular momentum phenomenon called precession.
b) It has a linear velocity relative to the tower, a linear momentum phenomenon, which I have given a working title of Dark Motion.
Point a) is fully documented, discussed and understood by classical physics.
Point b) is not and is responsible for the elusive and mysterious which we pursue.
The question Ram sent to “ask a physicist ” remains one of sciences greatest mysteries. Not what does it do in real life, that is simple to determine by experiment, oh no, the mystery is why it has not been measured, by every physics department in every University in the world!!!
Momentus
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Albert Druid - 16/08/2010 01:12:01
| | | Glad you're still on with gyros Momentus,
Like many others Nitro knew precession results from gravity acting on momentum of the flywheel - he had to be lookin for deeper reasoning. Your explanation brings deeper light and also more questions. I think Glenn said gravity in freefall builds momentum a little at a time - i like that cause then i can compare momentum built by gravity acting on momentum of the flywheel - apples to apples - I hope that's what Glenn was tryin to say - is what i got out of it after many a reading.
A believe you're right Momentus - the mysterious side of gyros is the motion outside c of g. The maths look simple and work real good when it comes to precession speed even for motion outside c of g - except i cant find no maths on how it get from stand still to precession -its all gummed up by the instantaneous.
Thanks for continuing the effort.
Al Druid
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Glenn Hawkins - 17/08/2010 02:16:25
| | | Since you have a good mind and you are interested I will teach you as time permits. Let me say first, I know more because I have studied more, not necessarily, because I am smarter than anybody else is. To begin. . .
A baseball is thrown 100 miles per hour. As it travels from place of departure to place of arrival, a distance of 100 feet, it encounters a crosswind, also of 100 miles per hour. You can draw this on a sheet of paper and see the beginning path of the ball was 180 degrees (straight) but due to the side wind the ball curves to arrive at an angle greater than 90 degrees.
Redo this though experiment and draw again. Do it all exactly as before, except have Superman through the ball ant 1,000 MPH. You will find that the ball will be off course by only nine decrees plus a little.
The difference was that the Superman’s ball traveled so fast in the given distance that the side wind had only 1/10 of the time to act on the fastball than did the slower ball. The consequence of less time to apply pressure per measured distance brings us to the assumption that the faster ball resisted sideways force 10 times more than a slower moving ball. That is not the reason as you might deduce by this thought experiment, but it is how we all relate speed, time and resistance.
Having said that, the theory of relativity predicts that increasing speeds increases resistance. That in fact is why light speed is supposed to have a limited, beyond which it is not possible to accelerate.
From this explanation, I expect us to understand the condition that mater traveling at high speed and consequently increased momentum, resist force that would alter its course. The faster the matter travels the more it resist (actually, the less time it has to respond to pressure).
Each complete revolution of a wheel means that matter is traveling, though in a loop, still nevertheless at a straight line relative to the plane it rotates in. Each reevaluation can be converted by pi into a linear distance. From here we should be able to surmise why and how a rotating wheel resist being tilted any direction and that the greater the mass, the greater the diameter, the greater the speed of rotation, the greater the resistance.
Lastly, some of the most involved hands-on testing ever done was in 1940. I speak of the various experts from the several sciences that were brought together by the military to invent the first gyroscopic guidance system for American and British pilots who were crashing, flying in the fog during World War ll. Among the things they jointly ended up stating was, “A gyroscope without friction would precess indefinitely into the future. They meant, ‘nearly infinitely‘, but were cautious. Can you now, with this explanation see they were?
A gyroscope cannot precess unless it first tiltes. It cannot be tilted by gravity, without first falling into gravity. Yes, the faster it rotates the slower it precesses and the slower it fall, but fall it must.
So which drives which? The fall must first occur in order to begin the precession. This is not a chicken and egg, which came first question. So, if the fall must comes first, which of course it dose, then the beginning of precession can not be simultaneous to the thing that causes it. Always the state of precession will be a reaction to the tilting fall into gravity which preceded it. To think otherwise is to think that when you push a heavy object by hand, it moves by itself, simultaneously, rather than by cause and effect, action and reaction. Always an increment of precession, will be the result of a past and prier increment movement -- the drop 'n' tilt.
This has been a mechanical explanation of why and how a spinning wheel resists being tilted. In addition, simultaneous action and reaction is not possible. There may be more explanation as I have time. There sure is a lot left untold here.
I am not going to waist my time proof reading this. I hope it makes sense and is not too screwed up. Please take what ever you can from it, as it is.
I want to compliment you very much, on your insight into the fact that one correct measurement is-- momentum to momentum as you deduced.
Best Regards,
Glenn
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Glenn Hawkins - 17/08/2010 02:26:56
| | | ". . . they jointly ended up stating, “A gyroscope without friction would precess indefinitely into the future -- .
I meant to say they were -- WRONG!
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Albert Druid - 19/08/2010 18:48:09
| | | You done it again Glenn - off with a great start - to focus on reduced total action time for the crosswind - it help me see the gyro picture more clear. Even if i mostly have difficulty with the instantaneous - am not on board with your notion that actions and reactions are not simultaneous - cause & effect, or action & reaction don't need happen in temporal sequence - the action need only deliver force whilst the reaction receives the effect of the force. I push or lift a bucket - my hand acts with a force - the bucket had no initial force - it gets momentum from the hand - both, hand and bucket, move simultaneous. Simultaneous ain't the same as it moves by itself - by itself motion don't require anything else to push it, simultaneous do.
In ma thinking precession is a gyro's natural response to torque, same way that falling is natural in linear motion - any gyro drop motion come in play when inertia of gyro mass resist start of precessing - resistance present a torque producing a new deflection which is normal gyro response to torque.
If precession need a prior increment of drop, before it start - what's the size of that inclement? - how's increment size determined? is there a minimum quantum of increment?
Thanks for both of the fine tools to help thinking bout gyros - 1 gravity motion accumulates momentum and 2 focus on time per measured distance.
Al Druid
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Glenn Hawskins - 20/08/2010 01:07:56
| | | Simultaneous, or none-simultaneous: the question has no useful meaning to the purpose and functions in seeking to build inertial propulsion. Our ideas on this condition are purely analytical play, though we all found it interesting.
I think I have beaten up enough on whether simultaneous is-or-isn’t. Though I disagree with you and Momentous, I admire your minds and now yield to you, for the nature of the question is pointless and so acquiescence is easy for me. I know how to produce propulsion. And whether simultaneous is, or whether force begins and ends as an action and reaction sequence as Isaac Newton said, doesn’t matter at all in the building of inertial propulsion. Both ideas work equally well.
I want to complement both of you for being able to burn your candles at high glow in the darkness of this area.
Take Care,
Glenn
I think you are both quick and sharp.
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Glenn Hawskins - 20/08/2010 01:08:03
| | | Simultaneous, or none-simultaneous: the question has no useful meaning to the purpose and functions in seeking to build inertial propulsion. Our ideas on this condition are purely analytical play, though we all found it interesting.
I think I have beaten up enough on whether simultaneous is-or-isn’t. Though I disagree with you and Momentous, I admire your minds and now yield to you, for the nature of the question is pointless and so acquiescence is easy for me. I know how to produce propulsion. And whether simultaneous is, or whether force begins and ends as an action and reaction sequence as Isaac Newton said, doesn’t matter at all in the building of inertial propulsion. Both ideas work equally well.
I want to complement both of you for being able to burn your candles at high glow in the darkness of this area.
Take Care,
Glenn
I think you are both quick and sharp.
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Albert Druid - 20/08/2010 14:52:22
| | | Thanks for the compliments Glenn -
this forum's great cause everyone can learn from everyone. it don't matter how we chose to present a picture - so long as it make the picture clearer in our mind. i think there's folks who understand the maths of gyros all the way through - they explanations is hard t figure for regular folks - a think even they don't fully understand the force flows enough to determine if propulsion machines are posible. See you round.
Al Druid
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Glenn Hawkins - 21/08/2010 01:03:22
| | | You say, “this forum's great cause everyone can learn from everyone.”
I’m afraid not Al. Nobody has ever given the slightest hint of worthwhile knowledge towards propulsion-- nothing, nothing whatsoever in all the years. They do not know. None ever knew. I am the only person in the world who knows.
Good Luck and goodbye for a while,
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Albert Druid - 23/08/2010 13:19:03
| | | Good luck Glenn - see you in the news when your famous.
Al Druid
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patrick - 31/08/2010 01:22:11
| | | ALBER
Of what drop do they state causes motion?
The whole entropic force of dropping thru a magnetic repultion device may be a drop or drip in time but the whole point of a circular linnear device is 2 take electrons laterally out using their own spin and massive repution and channel it through a gyro conicle cotton reel shaped larger at the base than top,with a carved spiggot body that winds around itself. AND also the same on the face of the inner face of the surrounding body of the external gyro.
Carved outer body 359, carved inner 367. Both pime numbers neither faces of carved spiggots would ever meet B.Cos( A the heliscopic faces dont alighn) (B two prime numbers cannot allighn in 360 degrees) (The drop only allows a pin point allignment but the line of before and after that point have two entirely different qualities.
POSSITRONS collected central and under repultion thrown down.
Only to meet new spun entricle in charged chamber. Control of BETA and ALFA DECAY.
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Mike - 07/09/2010 02:01:21
| | | here is a full page on the "Drop" explained
http://www.mb-soft.com/public3/gravit33.html
and
http://www.mb-soft.com/public/precess.html
and
"Keep in mind that the Soviet Union had spent more than 20 years and many millions of Rubles in attempting to build devices to achieve this goal."
http://www.mb-soft.com/public2/earthrot.html
lots of reading
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Albert Druid - 08/09/2010 18:51:36
| | | you all are way over my head - Al
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Glenn Hawkins - 09/09/2010 03:44:02
| | | Hi Al,
Sometime I will offer very simple observations and solutions that give you everything you need and want. You’ll have no problem.
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Albert Druid - 14/09/2010 15:38:19
| | | appreciate all your help - wishing my best - Al Druid
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Albert Druid - 20/09/2010 19:22:15
| | | only answers am interested is bout calculatin strengths of precession -
we know formulas to calculate
- linear momentum
- linear force
- angular momentum
- angular torque
But i aint seen no formulas to calculate momentum - force - torques - or any such similar fer precession
i hopes anyone smarter than me know the answer
thnx Al.
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Glenn Hawkins - 20/09/2010 20:14:46
| | | Well, what good will it do to know those answers? One could always go to school, but anyway, I think you may find those answers already given on this site. Those answers are child’s play to the world’s most gifted scholars. They’ve know how to do the math for those ‘answers’ for centuries and their math told them inertial propulsion is not possible.
I tell you it is possible and knowing ‘how to do it mechanically, not mathematically’ is the only worthwhile IP information there is at this time. The math has to follow the invention. Math is learned. Somebody taught somebody who taught somebody.
Original mechanical cognition in the mind, the search, the findings, the relentless learning and effort and then if your are lucky, occasional divine explosions in your cerebral from time to time until you finally have it all--- that is the primary means of invention. It always has been, though math is certainly a magnificent tool also used wonderfully by inventors-- however:
You are the one who invented fire by rubbing two stones together. After you succeeded when no one else could, they told you how fast to rub and how much pressure to apply and for a given time-- all mathematical products after the fact, not before the fact.
Your post are actually good, though I might not make that clear sometimes.
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Albert Druid - 21/09/2010 02:14:41
| | | its ok Glenn - someone else is bound to know the answer- like you say. fact is i aint found the equations here or anywhere - can't even find what book or school course to take.
i like maps when going somewhere new - maps don't always help but sometimes.
good thing our lord made us all different and each man finds things his way, else we may still be halfway from the stone age - anyway its all bout curiosity - am sure inventing takes a lot of diferent activities, like you say.
oh - do anyone know the calculation formula for time from zero to precession max speed? am lookin foward to your observations
Al Druid
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Luis Gonzalez - 25/09/2010 01:54:16
| | | Hi Al,
Thank you for the posting on the other thread and sorry for the slow response.
I believe I can be of help regarding some of your questions. I may not provide you with gospel truth but the equation I have derived provide good and simple approximations (in some cases they are exactly accurate).
I will start with the most readily accessible equation, which is perhaps not too difficult to derive. This is the equation to calculate the time increment required for the gyro to accelerate from zero to the maximum velocity of steady precession, in a standard gyroscope (this equation varies for other configurations).
The time increment can be calculated by taking the ratio of the axle's length to the flywheel's radius, and SQUARING this ratio, and then Dividing the SQUARED result by "W', the angular velocity of the flywheel.
The result is a very small number. For example, an average size gyro spinning at about 12,000 RPM will take about 1/800 to 1/1,200 of a second to go from zero to maximum steady precession velocity.
The most interesting part is that changing the weight of the flywheel or increasing the torque of gravity (if it were possible) does not affect this time increment at all! Only the flywheel's angular velocity and the ratio of the 2 radii make a difference!
I will let you think about these initial facts before going on to the next step, which is an equation to calculate the distance that the gyro drops.
Best Regards
Luis G
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Albert Druid - 28/09/2010 01:30:18
| | | appreciate your formula for time Luis - am eager for the distance formula. where can we find this kind of formulas? Thx Al Druid
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Luis Gonzalez - 01/10/2010 03:06:33
| | | You are welcome Al,
I am not aware of any place to find the equations you seek; they probably exist in a number of places buried among symbols that make them intelligible to laymen like us.
The equations I provide can be derived by any determined individual with an understanding of basic physics and a large dose of curiosity. I just derived them, that is why I mentioned that they are not to be taken as gospel truth, but rather as approximations (that is what science really is).
Once we know the acceleration behind torque and the time increment, calculating the distance of the "drop" is just a matter of applying a well established physics equation for distance = (1/2) * (A * t * t)
In other words, the distance is equal to half the acceleration, multiplied by time (t) SQUARED.
We already know the equation for the time (t) increment, so we SQUARE that equation and multiply it by 16, which is half the value of gravity's acceleration.
You may be thinking that since the equation for the time increment had variables raised to the second power (SQUARED) and now we need to SQUARE them one more time, we will take these variables to the FOURTH (4th) POWER.
That is correct!
Basic angular motion is often not entirely well understood:
For example the units of length for most basic "Angular" calculations are squared while the units of length for "Linear" calculation are NOT Squared (intuitively I think this occurs because centripetal acceleration is always present in all angular motions, just a guess).
I have never seen these equations in this forum, anywhere in the web, or in any of the physics books that I can get my hands on and understand. Perhaps I am not smart enough to understand the books that contain this type of information; in my opinion, the math has been greatly mystified by those who write it, and the broad intricacies of gyros have largely ignored or only been narrowly delved into.
Let me know when you are ready for my notion regarding the force of precession itself (which in my opinion is not as important as the equations for time-increment, and for distance). Or perhaps you will want to know how I derived the 2 equations above.
Best Regards,
Luis G
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Albert Druid - 01/10/2010 20:55:26
| | | thx again Luis - am ok figuring what you given me - can you tell me bout the strength of gyros? - you think is possible to get propulsion from gyros?
good wishes Al Druid
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Luis Gonzalez - 18/10/2010 02:02:18
| | | Good questions Al,
My apology for the long wait, I don't visit this forum as often as I used to.
Let's define your use of the word "Strength" (in your question) in regard to gyros. We will define gyro "Strength", as motions having a component of active Acceleration. If the "Acceleration" factor were to be replaced by a factor of "Velocity", then the motion would stops being a "Strength" and would become only a POTENTIAL "Strength", because when the Acceleration factor is removed, then the motion is NOT a Force, not a Torque, and not related to this family of motions.
On the other hand, a "Velocity" factor enables "momentum", which can only store (and can release) a potential and LIMITED amount of stored "Strength / Force / Torque / etc".
Interestingly, steady precession has a factor of "Velocity", but NO "Acceleration" factor in the direction of its motion. Precession's motion is analogous to the angular motion in a spinning object (or in a satellite) because both of these motions require the coexistence of a STATIC "Strength" (spin and satellites require centripetal Force, and precession requires a Torque). And in both cases the static force occurs perpendicular to the actual motion.
Therefore steady precession can only store a limited quantity of POTENTIAL "Strength". This strength becomes stored during the initial "drop" when precession's motion is "Accelerated" to its steady maximum velocity.
To fully answer your question, the total magnitude of the "Strength" stored in precession can be calculated by taking the product of precession's angular velocity times the radius of precession and SQUARING that amount, and then multiplying that result by the Mass...
i.e. (Wp) * (Rp) * (Wp) * (Rp) * (M).
This equation is provided as an approximation but is most likely exact.
I think the above statements complete my answer to the first of your last 2 questions.
On my next posting I will try to provide my best answer to your most important last question.
Meanwhile, perhaps other contributors would like to share their answer Al; your question was, is it possible to get propulsion by using the effects of gyros?
My Regards,
Luis G
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Albert Druid - 18/10/2010 14:01:39
| | | Luis - no need to apologize, appreciate your direct answers no games no power trips - wish other folks were willing to share what they know - Glenn was real helpful but I think he is busy building his invention
Thx Al
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Albert Druid - 20/10/2010 21:12:18
| | | Hey Luis - am I right'n thinkin that nobody else has thought through the information you given me? Al
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Harry K. - 24/10/2010 13:43:10
| | | Hello Albert,
No you're not right. If you reading through this forum you may find many explanations and and calculation maths. Luis has often very good ideas in gyro related issues but in my opinion he is sometimes on the wrong track regarding basic gyro behavior explanations. But he is a fine guy who has spent a lot of time and best effort in posting and explaning many interesting gyro related issues. Thanks, Luis!
However, it's very difficult to provide for others understandable explanations with such a complex stuff, and also if applicable in a foreign language (e.g. my native language is German).
Have a nice Sunday!
Harry
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Albert Druid - 27/10/2010 01:34:39
| | | Thx Harry - do you have better formulas than Luis to calculate - time span that gyros drops? distance gyros drops? the strength or force in gyros?
Al
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Harry K. - 30/10/2010 13:07:20
| | | Hello Albert,
Sorry for my late response but unfortunately I have less time in the moment.
Your initial question was:
"Nitro asked - why a gyro goes swiftly to a set speed when released - one might reasonably expect it to keep accelerating under the constant force of gravity? "
I guess we talking about an overhung bedded gyroscope which can precess around a hub?
In this case the following geometrical and physical parameters are involved:
1. Geometrical shape of the flywheel (disk, wheel, globe, sphere, etc.)
2. Mass of the flywheel: mG
3. Inertia of the flywheel around the center axis of rotating mass: mGx
4. Inertia of the flywheel around the center axis of precessing mass: mGy
5. Radius of flywheel: rG
6. Radius of precessing flywheel around the hub: rD
7. Gravitation acceleration: g
8. Angular velocity of spinning mass: (w means "Omega") wG
With these parameters you are able to calculate things like precession speed, tilting torque, kinetic and potential energy, centrifugal force.
However, at first you have to understand what really happens and why this happens. And here we are again at your initial question: "why a gyro goes swiftly to a set speed when released".
Precession occurs because a torque tries to rotate a spinning mass orthogonal to its spinning plane. In this case, the angular momentum of each spinning mass point gets an additional angular momentum vetor, which is acting orthogonal. This additional angular momentum vector is acting orthogonal because the tilting torque is acting orthogonal to the spinning plane of the flywheel mass.
The precession velocity is caused by the ratio of angular momentum vector of tilting torque and angular momentum vector of spinning mass. As long as both angular momentum vectors remain constant, the ratio between both vectors remain constant and thus precession velocity remain constant as well.
Anyway, the precession velocity of an overhung bedded gyroscope will not remain constant because of the following issues:
1. The mass of the spinning flywheel has to be accelerated from zero to precession speed.
2. As soon as the spinning flywheel (=gyroscope) begins to precess, a new torque, caused by the centrifugal force of the precessing flywheel, is acting against the tilting torque, caused by gravitation force acting at the flywheel.
3. Because of friction losses, caused by friction in bearings and air resistance, spinning and precession speed will decelerate.
Some people may have problems to imagine Issue 1. However, it's clear that energy is necessary to accelerate a mass to a specific velocity, isn't it?
This energy can be easily calculated with the parameters of precession speed and mass inertia of the flywheel in preccession plane.
This kinetic energy is equal to the potential energy, which can be calculated by gyro mass and gravity height of the dropped gyro.
If both equations for kinetic and potential energy are equalized, the dropping angle "Alpha" and also the dropping height can be calculated.
Angle "Alpha" or the dropping height will become greater, if:
- Mass inertia of the flywheel (in direction of hub plane) increases
- if precession speed increases
But on the other hand (issue 2) the dropping of the gyro will be decelerated if:
- the centrifugal force of the precessing gyro increases, i.e. if the precessing speed increases
- the dropping angle "Alpha" will become greater, because the tilting torque caused by gravity becomes smaller by factor cosinus "Alpha".
Issue 3 is responsible for the fact, that a gyroscope precesses faster and faster, because the spinning velocity becomes for friction reasons slower and slower and thus the ratio between tilting torque and angular momentum of the spinning mass becomes smaller, i.e. precession speed increases.
So you see, on the one hand there a standard and well-known physical equations to calculate all kind of gyroscopic related issues such as percession speed, dropping angle, centrifugal forces, etc. but on the other hand it's difficult to calculate some of these date exactly (dropping angle). I think it's only possible to calculate an approximation value of the really dropping angle "Alpha".
However, in my opinion the exactly calculated angle is not important at all, but it's important to understand what really happens.
Unfortunately, in my opinion it's not possible to achieve thrust with the help of gyroscopes, because all involved forces can only act at a gyroscope in form of torques. And a torque is defined by 2 opposite directed forces with a parallel distance to each other. That means, that both opposite directed, parallel aligned forces will always nullify each other.
I hope my statement is in some degree understandable for you. Comments from you or other contributors are appreciated.
Have a nice weekend!
Harry
@Glenn H.
Hello Glenn! - I have tried to use less maths as pssible. Hope, everything is fine with you!
Take care!
Harry
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Albert Druid - 04/11/2010 03:03:17
| | | Hi Harry - i read your fine explanations a number of times -- can't figure out what is your formula to calculate the time span that the gyro drop - and the distance the gyro drops - do you have these formulas? Thanks Al
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Glenn Hawkins - 05/11/2010 15:48:02
| | | @Harry K.
Hello Harry! - It’s good to hear from you. You do magnificent math I think, because not all of it is over my head, beyond judging. Math, as you know better than I, is the universal languish of science--not my mechanics. So plaster the stuff all over the board and I will be happy for you. You are my favorite German, next to the memory of a girl (Chris) in Friedberg so long, long ago. Then the history of Field Marshal Erwin Rommel comes in third just behind you. You then are only second. I am sorry. Just kidding you.
Take care of yourself and family,
Glenn
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fahkei@gmx.de - 07/11/2010 18:42:49
| | | Hi Albert,
The formula to calculate the time span that the gyro drop and the distance the gyro drops are basic phsical equations as I have mentioned in my last post.
I don't about your education in physics and maths, but if you want I can send you the involved formula by email. Contrary to me you should know my email address, so you can send me a short message.
Regards,
Harry
Hello Glenn,
Thank you for your nice words!
Take care, too!
Harry
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Albert Druid - 08/11/2010 01:31:10
| | | Hi Harry - i have no physics education except books friends and the now the internet - am glad these formulas are basic so most folks with education should know them - would you mind posting them and sharing them with us in this forum?
i frankly don't wish to share my daughter in law's email with any one - i don't have much to do with computers and don't have inclination to spend time and money on one - appreciate your help
thanks Al
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Harry K. - 09/11/2010 20:19:17
| | | Hi Albert,
I understand that you will not share your daughter in law's email. However, posted equations here in this forum are difficult to read und thus not easy to understand.
I have already posted a many links to my server regarding calculation equations, sketches, etc and equations in my balance theory thread:
http://www.gyroscopes.org/forum/questions.asp?id=922
(Copy and paste this link into the URL window of your browser)
Here are 2 links regarding calculation of geometrical shape of spinning masses and a formular sheet to calculate kinetic energy for precession accleration and drop of axis in tilting direction (I think this is what you are looking for):
http://www.misc.keipert.net/gyro/Mass_Inertia.pdf
http://www.misc.keipert.net/gyro/gyro-calculations-1.pdf
I will prepare another formula sheet to calculate kinetic energy and drop of axis for an overhung gyroscope, but this will take some time because I'm very busy at the moment. However, the equations are mainly the same, only centripetal forces will be added.
I hope these equations are understandable for you. If not please let me know and I will try to explain.
Regards,
Harry
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Sandy Kidd - 09/11/2010 21:25:06
| | | Albert Druid, Shed Dwellers and other interested parties,
With some interest I read replies offered up to Nitro’s now quite dated statement referring to precession movement in a gravity accelerated gyroscope system.
Having followed the input to this site for some long time, and including the original publication of Nitro’s statement, I am still keen to understand why some people believe that there is something relating to gyroscopic action which will provide the key to inertial drive.
Certainly gyroscopes can provide some very strange reactions when subjected to certain forces, but is this in itself enough to make one believe in the possible creation of inertial thrust, and sizeable lumps of it, at that?
Glenn Hawkins’ last remarks prompted this reply.
Seems that the “Shed Dwellers” are fairly well divided in which path they are prepared to follow in pursuit of answers.
There seems to be 3 pretty well defined directions of pursuit favoured by the “Shed Dwellers” if I may call you that.
Route 1
It would appear, certainly to me, that the present 2 or 3 decade “craze” was originally started by practical experiments designed and carried out by one Professor Eric Laithwaite which he claimed showed that gyroscopes could defy the accepted rules of motion, i.e. Newton’s Laws. His claims as everyone knows, destroyed him professionally. He was ridiculed, but much of his claim was seized upon by like-minded others who believed that he was not so wrong and that answers were required for much of his claims
The point is that these people believe that there was something not quite right with physics, they did not know what it was but many of them were prepared to give it a try and hoped to find out what.
I think it would be fair to say that the practical inventors, builders etc belong in this group.
Route 2
As a result of the fact that there was a prior presence in the use of gyroscopes to create inertial thrust, many others have got themselves involved in this game if I may call it that.
Most have been very selective in what they wish to believe, and in most cases, their belief is based on some personal notion, fad, or fancy which will be supported for no good reason, other than the fact, that they have chosen it.
I believe that the bulk of this squad have not really thought of the implications of what they subscribe to, whether the Laws of Motion need to be breached or not.
Admittedly over the years a great deal of very interesting information, (correct or otherwise) has been published on this site, but in most cases the flavour of the information has not suited other readers’ tastes.
The information, admittedly at no cost to the reader, has therefore been rejected.
Irrespective of the arguments presented by a few of the subscribers there have been very few agreements over the last many years.
The problem here is that information which I personally think was worthy of further investigation has very often been drowned by junk, well meaning and otherwise.
One of our most prolific contributors suggests that it has all been a waste of time, with no contributions of any worth being offered and nothing has been learned or gained.
At this time this would appear to be true.
Route 3
In this section we have as Glenn Turner suggests the mathematician, physicist and engineer battalion.
They by their professional status must surely be obliged to accept the Laws of Motion as if they were cast in stone and are searching; it seems to me, for an anomaly in the interpretation of the laws which would allow for the production of inertial thrust.
We are assured that there is no such anomaly although it could be very interesting (if suitably educated) to manipulate the math and see what transpires.
The only cost is in time and at least it does not involve the expensive and time consuming manufacture, of in most cases, useless test devices.
I agree Albert the passion has gone and a deafening silence now exists.
Has everyone at last run out of options?
There is no magic in the gyroscope?
“Uninvolved but Still Interested”
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Albert Druid - 11/11/2010 02:54:01
| | | Hi Sandy your opinion is always welcome - thanks for presenting your experiments in this site - as i said earlier its fortunate people think different ways - else we would all be building fires the way it was done 5 thousand years ago. wish you well - Al
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Glenn Hawkins - 11/11/2010 17:04:28
| | | Sandy, “I am still keen to understand why some people believe that there is something relating to gyroscopic action which will provide the key to inertial drive.”
Glenn, ‘ Kind of funny Sandy, due to the fact that you know these forces’ and you know they can be used for propulsion. Anyway, you’ve spurned me to answer about why I know, more than how I know. Your information is forthcoming in a new thread above.
Sandy, “. . . these people believe that there was something not quite right with physics,”
Glenn, ‘I know you've learned as I have that it is impossible to cause a man to question what he learned as a child. Physics’ sole purpose is to measure and from that process learned the unknown to a such a great degree that is almost unbelievable. However, I have a few examples that show where physics has invented wrong explanation of how actual mechanical truths work in order to allow the correct measuring of those things. I approve happily. They did the right thing. Still, there are truths masked over that you are so very much aware of. Of course your implied thinking is right.
Sandy, “The problem here is that information which I personally think was worthy of further investigation has very often been drowned by junk, well meaning and otherwise.
One of our most prolific contributors suggests that it has all been a waste of time, with no contributions of any worth being offered and nothing has been learned or gained.
At this time this would appear to be true.”
Glenn, ‘I hope not to offends anyone, but yes to that Sandy. The electric motor was invented by a blacksmith in Philadelphia after hearing Benjamin Franklin espouse on then little know magnetism. I doubt the smitty had any education. Now days however, if you look at micro motors you think these modern engineered motor came from advanced civilizations in outer space they are so magnificent and advanced. However, every advancement of them was an invention. How to do it better with exactitude relied on engineering and physics-- but always first there had be an additional invention idea. Well done bozs! But who has invented powerful inertial thrust? No one. . . ! so the invention must come first before it can be mathamaticalized and engineered. This is all backwards bozs. This forum is titled ‘Gyro Propulsion’. The horse is harnessed backwards lately. First must come the invention. Then everyone can measure it and make it better. Right now measuring is an exercise and not a possible means. Invention is an idea. Measuring is a process. Lately there has not been much to process around here.
Sandy, “I agree Albert the passion has gone and a deafening silence now exists.
Has everyone at last run out of options?”
Glenn, “I’m sorry Sandy. Yes that’s right. I’m beginning to spill the beans with this new thread I‘ve just added. My experience is that no one will be interested, but I’ll do it anyway and leave it for posterity. In case somebody want’s to build it, they‘ll know how.
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Glenn Hawkins - 12/11/2010 02:39:32
| | | Hi Harry,
I read your post and enjoyed it. I understand it far easer than in the past and think it is fine. I’ll comment on a couple of things.
Harry, “Angle "Alpha" or the dropping height will become greater, if:
- Mass inertia of the flywheel (in direction of hub plane) increases
- if precession speed increases”
Glenn [[[ I don’t believe for a minute you have any understanding wrong, but maybe the explanation is confusing? The ‘dropping height’ is set at the time of release. It will not become greater, but lesser for as the gyro tilts, it will become lower-- which it always must do with, or without friction. The drop is a secondary power source to continue the action. Don’t you mean (I’m sure you do) that the dropping speed will be faster if any of the several condition that are possible, causes the precession speed to increase? Funny, I know you know, but still. . . ]]]
Harry, “But on the other hand (issue 2) the dropping of the gyro will be decelerated if:
- the centrifugal force of the precessing gyro increases,. . .“
Glenn [[[ Do I understand? Do you mean the centrifuge in the spinning wheel, or the centrifuge in the revolving precession? Yes of course to this-- the faster the wheel is rotated, the more centrifuge, the slower the drop-- but what would cause these increases in centrifuge? They‘d have to be manipulated to accelerate while precessiong. ]]]
Harry, “But on the other hand (issue 2) the dropping of the gyro will be decelerated if:. . . i.e. if the precessing speed increases.”
Glenn [[[ Sorry, but if the dropping speed -accelerates- (not decelerate) the precesion speed will conversely increase.]]]
Glenn [[[There is no question but that I know you understand perfectly. I hope my suggestions will help you arranges your explanations.]]]
Good Regards and good intentions,
Glenn
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Glenn Hawkins - 12/11/2010 16:25:56
| | | Hello Albert,
YOUR POST IS ABOUT. . .
Nitro asked - “. . . why a gyro goes swiftly to a set speed when released - one might reasonably expect it to keep accelerating under the constant force of gravity?”
Here are exact answers:
First: The gyro goes swiftly to a set speed, because all the interlocking forces that support the gyro from falling are in play in the immediate. They are not in hesitation, but are active in microseconds. Look to the speed of rotating particle in the wheel. That is how fast the transition of rotation into support travels. I once calculated I could accelerate a Taco gyroscope with a long string to 27,000 RPMs. That is, each particle makes a complete rotation every 450 th. of a second = 450/1 sec. That is almost how fast the gyro reaches ‘set speed’ as you are calling it. Can you perceive that the rotation is the support, which is resisting deflection towards the drop and that the support is in play before the droop ever begins? If my explanation is well done--, then you understand the answer completely.
Second: Why doesn’t a gyro keep accelerating under the constant force of gravity? There is a force of resistance. Think of a bow and arrow. The further you pull the bow the harder it becomes to pull. In a gyro, the partials in rotation resist being altered from their plain and circular path. They wish to rotate in their circular path without being deflected, which is what happens when the gyro drops into tilt. Like all ‘levers’ you studied in physics, time, speed, mass and distance determined how much force is required to lift a rock with a pry bar. Also, remember tackles and chains and the same holds true. Now then, a gyro must tilt, as you say fall, even if it is only 1 mm per year, otherwise there is no precession possible. The fall then is controlled by the unwillingness of the rotating partials to change their course in a hurry. If the wheel tried to fall faster than the speed it should fall, considering all the conditions, then the rotating particle resistance would become greater, until a balance was reached. In summation, a gyroscope cannot fall faster than the rotating particles are willing to yield from their spin path.
Regards,
Glenn
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Albert Druid - 13/11/2010 00:30:02
| | | thank for the visual explanations Glenn - believe it or not the previous answers answered these questions - now am interested in a basic formula to calculate the drop time - thanks again Al
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Glenn Hawkins - 13/11/2010 00:34:36
| | | Hi All,
But I am sure they did not. You explain how they did.
Glenn
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Albert Druid - 13/11/2010 15:40:17
| | | am no writer - even all who know how to explained do it with difficulty - for me is enough to understand that angular momentum and torque achieve balance - same ways liner momentum and force achieve balance - like happens in satellites - Al
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Glenn Hawkins - 13/11/2010 17:16:42
| | |
There is no damn drop time. The decent acceleration is governed instantaneously, owing to the degree of resistance that existed prior to the gyro ever being let go. A drop time never existed. You are confusing the gradual yielding of deflections allowing for tilting. There is no acceleration, no timing, no drop time, only an alternating velocity governed by all the complex things Harry explained to you.
I think this is way over your head. There are brilliant people here who have studied gyroscopes for many years and learned things no one ever knew. You flippantly disregard their work, pretending to be competent to evaluate the work of your betters. It is a game, but come on. You don’t know chit. You have not done one bit of work. Neither have you responded with a hint of detail to the many details posted here to you.
This is a site for innovation and invention. Harry dose math, but it is very, very mechanical, beautifully mechanical. If you don’t like mechanics, at least buy a math book and study it before you give opinions.
There is no damn drop time. There is only a yielding resistance waiting to be tilted to prove it was always there. The speed of the yielding depends on all the factors involving torque, verses counter forces, which again, Harry explained. That’s all we can do for you. Buy the book.
By the way:
You say “angular momentum and torque achieve balance. . . “ Yes they do!
You say,” . . . same ways liner momentum and force achieve balance - like happens in satellites.” Sorry, but no.
The break-a-way momentum of an orbiting satellite, verses the pull of gravity has nothing to do with precession and what you want to call a ‘drop’, which was the point of this entire thread. Orbit is primarily only an example of rotation.
Confused?
//////////////////////////////////////////////////////////////
Look at the detailed body of work you got:
Glenn Hawkins - 12/11/2010 16:25:56
Hello Albert,
YOUR POST IS ABOUT. . .
Nitro asked - “. . . why a gyro goes swiftly to a set speed when released - one might reasonably expect it to keep accelerating under the constant force of gravity?”
Here are exact answers!
Look what your returned:
Albert Druid - 13/11/2010 00:30:02
“. . . believe it or not the previous answers answered these questions,”
“. . . now am interested in a basic formula to calculate the drop time,”
- thanks again Al
//////////////////////////////////////////////////////////////
Glenn Hawkins - 13/11/2010 00:34:36
Hi All,
But I am sure they did not. You explain how they did.
Glenn
Albert Druid - 13/11/2010 15:40:17
am no writer - bla, bla, bla, - Al
////////////////////////////////////////////////////////////////
Now then. This post will blow-over and be buried and forgotten in time like all the post before it. Learn good matters from it please, but otherwise forget this. Everyone here seems to like you, or at least not dislike you so don’t worry, just learn to be genuinely polite at heart, or it’s better to be silent.
Best Regards to you Albert,
Glenn
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Albert Druid - 13/11/2010 18:55:13
| | | didn't mean to upset anyone with my questions - i came to ask more than one question and am moving on to next question as the first one was answered to my own personal satisfaction - by drop time i mean the duration from zero to max precession time, which can't be instantaneous - i got math and physics books but none give this formula - sorry Glenn, as you own this site and can tell people what to ask and not, why don't you just delete what you think is useless - i like your website very much - Al
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Luis Gonzalez - 13/11/2010 20:01:16
| | | Hi Al,
I hope all is well as you continue your search for reliable equations in this subject.
The equations I gave you previously for "drop distance" and for strength/force/torque of precession are indeed missing the angle of elevation component, which Harry introduced in his response. You will need to factor in the appropriate trig function.
*** As stated before the "drop time" i.e. time increment from zero to maximum steady velocity of precession is obtained by SQUARING the ratio of the hub system's radius (gyro axel) to the gyro radius [ i.e. (Rs / Rg) * (Rs / Rg) ], and dividing that result by the angular velocity of the flywheel (i.e. the angular velocity of spin). ***
Harry also mentioned other peripheral variables such as friction etc, which can play a very large role when we try to observe the very small scales of energy involved in small toy gyros under gravity's torque.
Though the "drop" time increment is virtually independent of the gyro's torque and mass, the distance of the "drop" and forces involved are definitely affected by these mass and torque factors. Finally, the effects of peripheral factors such as friction etc may be reduced through appropriate gyro construction and configurations.
There are many opinions in this subject matter and this forum, and consensus is rarely found even among those who have been at it a long time, including among experts. Keep your curiosity active and enjoy the struggle of the quest.
I will try to have some semblance of an answer to your most important question which asks whether it is possible to get propulsion by using the effects of gyros.
By the way the name of the web site owner is "Glenn Turner".
Best Regards,
Luis G
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Sandy Kidd - 13/11/2010 20:44:27
| | | Shed Dwellers and other interested parties,
Luis managed to correct my error before I could.
My humble apologies to all, for getting our two Glenns mixed up.
It was of course Glen Hawkins I was referring to, and not our illustrious web master; Glenn Turner.
Old age is getting the better of me.
However can I add that whether the action/reaction that is causing all this heat is instantaneous or otherwise, it is my opinion (and only mine) that the answer, is not going to help anyone, anyway.
Do you have some idea, or plan in your head that requires a precise answer to this question or is it just interest.
Regards,
Sandy.
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Albert Druid - 13/11/2010 22:05:48
| | | hey Luis - thanks for clearing a couple of things - also isn't there balance between linear momentum in satellite and the force of gravity? if the velocity of a satellite changes it can fall or escape gravity - satellites in orbit look in balance to me.
it looks like maybe nobody else has a easy formula to figur the duration from zero to max precession velocity.
thanks for the posting Sandy - curiosity is my initial drive - i don't write well but i read and learn just fine.
to your question Sandy: - in my thinking the span of time that the gyro accelerates to max precession velocity happens before the saturation zone - mass exists before max precession velocity, and there is no mass to accelerate after max precession velocity - you see my interest?
Sandy i don't expect to produce inertial propulsion with gyros - i am satisfying my curiosity after reading through this site.
Glenn - i know angular motions are different than linear motions - i also understand some similarities like relationships between force and linear momentum resemble relations between torque and angular momentum - i know satellites dn't have precession - leastwise when they have no spin - my good wishes to all - Al
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Glenn Hawkins - 13/11/2010 22:41:21
| | |
Dear Albert,
There is a condition called nutation. You might enjoy looking it up on youtube for gyroscopes. It might be there. I have seen it get pretty, pretty wild. Also Wikipedia has information relating to satellite nutation.
Always a constant imbalance is present in all the vast number of objects in space and probably even down to quirks, near infinity smaller than atoms, and probably all shapes, actions and sizes in between i.e., gyroscopes wobble constantly in and out of balance. Sometimes we can see it greatly. Sometimes we can‘t see it happening with the necked eye. So when Harry indicated exactitude in some measurements was not necessary, I just smiled and though, Oh yeah, right you are.
By the way, I see you understand much more than is evident. Very good, Al.
Good hunting and well wishes,
Glenn
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Albert Druid - 14/11/2010 14:14:55
| | | Hi Harry - thanks for the many formulas - sorry i don't see a formula for the time duration from zero to max precession velocity - its possible am too dumb to see it - can you point it out for me? Al
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Luis Gonzalez - 14/11/2010 14:45:41
| | | Hi Al,
I am sure a few others must have thought about these equations. Harry and Ravi can certainly figure out how to derive them (though I think Harry may find something wrong in the way I presented them). With rare exceptions, I am pretty sure very few others in this forum know or understands these equations.
The most interesting thing Al, is that most likely nobody anywhere (except perhaps you) understands that these type of equations can be used to determine whether a given configuration can produce sustainable inertial propulsion or not!
Based on your response to Sandy, I was pleasantly surprised to find that you have grasped the gist of what I have zeroed-in..., that analyzing the increment from zero to full precession velocity holds the key to discovering sustainable inertial propulsion (not meaning to say that gravity gyros can ever produce sustainable inertial propulsion).
Sandy has been trying to tell this forum that once precession kicks in, it is not possible to extract propulsion with mechanically accelerated devices. And I have been touting that the acceleration of precession, through the tiny drop-increment, is where we can find the dynamics behind sustainable inertial propulsion. You grasped the connection between these "facts" and stated it more clearly than I or most anyone in this forum can.
I will allow time for others to chime in on your questions before posting my response to your question regarding the feasibility of sustainable inertial propulsion.
Best Regards,
Luis G
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Sandy Kidd - 14/11/2010 20:06:14
| | | Dear Luis,
If I may I would like to clarify your statement:
“Sandy has been trying to tell this forum that once precession kicks in, it is not possible to extract propulsion with mechanically accelerated devices”
That I suppose is true up to a point but worthy of a bit of clarification.
Precession is a normally seen in gravitational accelerated system in decay and as such does not exist in a mechanically accelerated system which is in effect a system full to overflowing , at what is being called precession.
Call it what you like but it is not precession.
Prior to this event copious amounts of thrust can be extracted, depending on the design of the system. Everything depends on the position of this critical point at which the effect of angular momentum on the system disappears.
In any gyroscope system which is rotating at a steady fixed speed it will be found that angular momentum and centrifugal force are at a maximum when gyroscope rotation speed is zero.
At some point when the gyroscopes are rotated sufficiently fast it will be found that the angular momentum and centrifugal force have been reduced to a minimum, in fact they disappear at the point the gyroscope is seen to ascend instead of continuing to rotate on a nearly flat plane. This is pretty obvious really as the gyroscope becomes virtually devoid of mass and its torque is all that is left.
In any gyroscope system it will be found that the angular momentum produced is proportional to the rotation speed of the gyroscope.
Consequently one does not design a device to run too close to that critical point in any gyroscope system.
There is oodles of angular momentum available for conversion well below that critical point, which can be designed to be at any level you want it to be.
Hope this serves to clarify my position,
Regards,
Sandy
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Luis Gonzalez - 14/11/2010 20:55:19
| | | Dear Sandy,
Thank you for the clarification.
Best Regards,
Luis G
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Glenn Hawkins - 15/11/2010 21:59:37
| | |
Come on guys! I’m trying to be kind.
What the hell dose it matter if precession acceleration time is 1,000/1 microseconds or I, 000,000/1 microseconds? It is simply near immediate, period.
Nitro Metcalf believed precession was instantaneous. That is fictitiously CORRECT! The meaning of fictitious is: ‘Accepted or assumed for the sake of convention!’ For hundreds of years the word and idea of fictitious, as most of us know well, was used to explain some conditions of rotation and the same word and explanation has been continued by every mathematician since. Instantaneous though not possible, is so close it puts everything in perspective and let’s us be done with it.
Anyway, unless someone can explained how a more exact answer could be of any use in the creation of propulsion, then it is ‘Junk” as Sandy hinted. (I think it is not possible anyway, because nothing is stable and nothing exactly repeats.)
The post started out OK. Nitro was referred to as asking ‘why‘-- not how much. I find that Harry and I both answered the two questions with different languish, but then the post changed to, I think, nonsensically. The new question is annoying and distractive even to know it is here. I move it be moved, and continued if you must, to the proper discussion threads provided here. Go to Main Forum Page. Choose, either ‘Basic Gyro Questions’ or choose ‘Physics + technical + maths of Gyroscopes‘. This section is for propulsion and relative math.
Important math questions here are many. Lots have been answered. The most important one I think, is ‘what are the mathematical parameters, parts and variations in construction that control the velocity of precession? Knowing how to build for precession velocity and precession circumference and such things is important to the attempt. They are fundamental.
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Harry K. - 15/11/2010 22:24:55
| | | Hello Albert & all,
Of course you are not too dumb to see this equation - quite the contrary! ;-)
The equation to calculate from zero to max precession velocity can be derived by the presented equations stated in the calculation sheet and some logical way of thinking:
1. In the sketch of the calculation sheet you can find the stated tilting torque "T", which can be also defined by the force couple "Ft" with the distance "rt" from the center of hub. In this sketch you can also imagine an acting dead weight mass on one of the gyro's axle. This dead weight mass with its distance to the cener of hub would also create a tilting torque "T". And an bedded gyroscope would create this tilting torque by its own dead weight mass. This is the only difference between both gyro systems, but the calculations are same foe both layouts.
2. As stated in my post from 30/10/2010, the mass of the spinning flywheel has to be accelerated from zero to precession speed. This process needs a certain time "t" to accelertate the mass of the spinning wheel an the dead weight mass in the plane of precession velocity. This time "t" depends on the precession velocity:
- the faster the precession velocity, the longer the acceleration time "t"
The reason for this fact is, that more ener´gy will be required for higher precession velocity.
3. Unfortunately I did not stated the droping distance "hdrop" and the droping angle "alpha" in the sketch of the calculation sheet, however, I hope you can imagine this both parameters.
tan(alpha) can be calculated by "hdrop" / rt
With the calculated precession energy "Ep" and in addition the required energy to accelerate the dead weight mass (if existent) you are able to calculate the time "t" from zero to precession velocity by this equation:
Ep (dead weight mass) = ( m (dead weight mass) / 2 ) * rt * rt * wp * wp
t = ( ARCTAN((Ep + Ep (dead weight mass) / T) ) / wp
(Ep - precession energy
Ep (dead weight mass) - dead weight mass which has to be accelerated in precession plane
T - tilting torque
wp - angular velocity of precession
entity . s (second))
Please note, that centrifugal forces, caused by dead weight masses, and friction forces are unconsidered!
As promised, I will post a calculation sheet for an overhung gyro system and with all required parameters in the sketch. And I will also show the complete derivation vor calculating the time increment, but please be patient.
I`m also curious for what reason you need this (very short) time span?
Best regards,
Harry
@Sandy
It`s nice to see that you still reading here. Your opinion regarding maths is well known but I hope you are broad-minded to post some calculations if somebody asks for something like that. If you mean with "saturation zone" the small time to accelerate from zero to precession speed then I will support you. - But only in this case... ;-)
Take care,
Harry
@Glenn
I will answer your statements later.
Best regards,
Harry
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Albert Druid - 16/11/2010 02:13:22
| | | thank you Harry - your established formulas are food for thought - i will need some time to digest and i look forward your calculation sheet - my interest in the drop time formula is to search for all possible ways to expand the time span - regards Al
i will not be able to post as often for some time
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Sandy Kidd - 17/11/2010 20:18:54
| | | Dear Harry,
For me to get involved in mathematics would prove nothing unless you accept my claims unconditionally and this is not going to happen is it?
Your physics and my physics are totally at odds because we believe in different reasons for the same thing.
Since 2004 or thereabouts I have dispensed enough information on this site for anyone with a bit of engineering in his soul to develop inertial thrust.
The amount of thrust depends a bit, on how good an engineer he is and if he has just a little bit of spatial aptitude.
Access to a little money would also help.
Luckily a lot of the latest model equipment has made this easy.
I can acquire the luxury of radio control with telemetry (pretty recent), at a reasonable price.
I can also acquire, very powerful, cheap, brushless outrunner motors,( they are effectively 3 phase), of excellent quality from China and programmable electronic speed controllers, also from China.
I had occasion to tell our Glenn about this stuff some time ago, as outrunner motors were what he was searching for.
I sourced all my stuff from an excellent company called “Giantcod”.
Their prices and delivery are second to none.
Try “Giantcod.co.uk.”
Have a look and see, if you are interested.
I wish it had been this simple 30 years ago.
However the rules are the same and the device will produce thrust for the same reasons, albeit much easier.
Harry a large leap of faith is required to get to grips with this thing, the physics being somewhat different from what you are used to.
Then your maths will come into their own.
As a matter of interest I have acquired all the necessary bits and pieces to assemble a device capable of gaining acceptance by the system.
I have a device well on, as far as completion is concerned.
With this device I was aiming to launch a full frontal attack.
By this I mean a public demonstration of vertical thrust of over 1G.
That makes it a bit harder to deny.
I have experienced the “Ostrich Syndrome” too many times before.
Unfortunately some time ago I suffered a rather severe heart attack and have incurred considerable damage to my heart. My time has been spent trying to recover from that, and my recovery has had to take priority.
Very soon though I hope to return to assemble my bits of kit..
Regards,
Sandy.
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Glenn Hawkins - 18/11/2010 01:52:25
| | | Yes Sandy,
You explained the outrunner motors and I am appreciative to you, to the extreme! I tried very heard to create better ones, because as you said the bearing construction is limited. I ran into all! kinds! of trouble. It seemed to me that government was set against me. It seems they knew what I wanted to do and didn’t want me to do it. Maybe I’m wrong and maybe stupid, but that is what I believe.
Regards Glenn,
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Glenn Hawkins - 18/11/2010 01:52:31
| | | Yes Sandy,
You explained the outrunner motors and I am appreciative to you, to the extreme! I tried very heard to create better ones, because as you said the bearing construction is limited. I ran into all! kinds! of trouble. It seemed to me that government was set against me. It seems they knew what I wanted to do and didn’t want me to do it. Maybe I’m wrong and maybe stupid, but that is what I believe.
Regards Glenn,
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Sandy Kidd - 18/11/2010 20:18:53
| | | Hello Glenn,
My sympathies.
If we could get someone to produce a brushless outrunner with decent bearings suitable for our purposes, a light, tidy, compact, device of considerable power could be made, but because of the constraints imposed by the light bearings, in my case my device will be much bigger than I wanted,
I was tempted to give them a try but it could be a lot of effort for little or no return.
Being a coward I went for the safe option, tried and tested by yours truly over many years.
I explained I intend to drive the discs through a universally coupled shaft to remove all loading from the motor bearings.
The control for the device rotation and each disc rotation on any one unit will not be any different because of this, it just makes the whole device a bit bigger than I wanted it to be. Not much heavier, just bigger.
One unit, my test unit, was well on when things turned nasty.
I will have to see how much effort the boss will let me put into it, she watches me like a hawk.
Regards,
Sandy
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patrick - 19/11/2010 07:47:48
| | | Torque is all that is left,there is no formula yet....BUT many minds Will make light work of that when forsight and aknolagement precludes!
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Glenn Hawkins - 19/11/2010 15:54:27
| | | Dear Sandy,
I was pleased to hear from you after so long. I entirely agree with you. Also, big is not always better, huh? Maybe in a Turkish bathhouse? I would not know, but I do know it is not better in our line of pursuit. Big can be bad and make everything too difficult and expensive for a small shop or rather, a guarder shed.
I have had the plains for cutting and shaping the design or sometime and the list of suppliers for several pre-made elements. A new kind of shaft must be secured or built. It must be larger, longer and hollow. The three insulated wired must run through the hollow shaft and connect to the stators. Then a good bearings system can be had.
[For the novice: stator is the electromagnets, in this case oppositely designed and of the more excellent design of the outrunner motor over the conventional motor. The electromagnets stators are connected to the shaft which rotates. When the shaft is locked down, the motor housing is force to rotate. This way the flywheel can be connected to the motor housing and spun. It is a wonderful idea and to me a critical idea. If you are here, because you are interested in I. P, then thank Sandy for the beginning insight.
My motor design dose not seem especially difficult to do, but it was damned difficult to learn how to do it.
I am sorry and disappointed that your latest effort is a problem. Still, you have started something totally new. This is a new development on this site. I wonder if any people here would catch on and seek to understand. There was once another old inertial propulsion site for engineers and when a new idea came out in it, it exploded with hundreds of new poster and lasted a couple of years before the thing was tested and failed.
Be happy you have a boss and hope that you cross the river before her. The one remaining dies many times.
Happiness to you an all yours,
Glenn
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Glenn Hawkins - 19/11/2010 15:57:31
| | | Dear Patrick,
You say, “Torque is all that is left. . .”
Sometimes you amaze me. I wonder why nobody else sees it. You could be exactly right.
You say, “. . . there is no formula yet. . .”
Well actually I think there is, but it hasn’t been simplified yet. Occam's Razor says that the simplest answered is the correct answer. All physicist and mathematicians use that simple premise. It causes Einstein to have to rewrite his Theory of Relativity.
Harry has most of the methods I believe, but they are long and tiring. Procedures and equations don’t give you the answer. They give you the method of finding the answer.
You did well, Pat.
Happy Holidays’
Glenn
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Albert Druid - 24/11/2010 01:55:05
| | | hi Harry - i got a handle on most of your formulas - except i can't make heads or tails on the formula for time span from zero to max precession velocity - this just don't look like normal formulas. i hope your calculation sheet will clear it up - thanks Al
happy thanksgiving to everyone
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patrick - 04/12/2010 01:12:59
| | | Primes 1
3
5
7
11
13
17
Then work back with difference between
1
2
0 3
0 2
0 0 5
0 2
2 7
0 4
2 11
2
See anything yet?
Look at the diagonals as well as thinking
1 divided by the lower which is 3 =0.3333
3 6
and so on....addition of the diagonals and toying with the inbetweens generates entropicity
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Luis Gonzalez - 11/12/2010 23:33:09
| | | Hi Al,
I don’t really expect that anyone is going to come up with an equation that makes sense for the time span that precession accelerates from zero to its maximum steady velocity.
It is true that you have to multiply my equation by Cos(alpha) if you want to drop the gyro at any angle other than horizontal. This is necessary to adjust for the true effective radius of the system, it is a minor adjustment (and I did state that my equations are intended to provide approximations).
Don’t hold your breath waiting for anyone else’s equation regarding the time span.
Perhaps (if you wish) some time we can kick around how to derive coherent gyro equations using the math that is already established publicly.
I am going to wait and see if anyone else is willing to volunteer answer with explanation to your ultimate question, whether inertial propulsion is indeed possible or not. Glenn has already started his own explanations in another thread.
Good luck and Merry Christmas to all.
Best Regards,
Luis G
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Albert Druid - 13/12/2010 02:15:48
| | | hey Luis – good to hear from you – thanks for the advice but am curious to see what others may say bout the time span formula – I expect all correct formulas will boil down to what formula you provided, but we never know. merry christmas – Al
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