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29 November 2024 03:38
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Question |
Asked by: |
Glenn Hawkins |
Subject: |
HELP ME OUT. |
Question: |
I need some assessments to help calibrate my design.
In overhung configurations I think the rate of precession relates to the rate of decent. You see this when the gyro is spun down in RPMs. Then the gyro precesses faster as it falls faster. If the speed is slowed just right, the motion is s 45 degrees angle between precession and decent.
(A.) What happens if you apply twenty times more mechanically force downward than gravity? Wouldn’t you have to multiply twenty times more RPMs than those slow ones which produced a 45 degrease decent under gravity?
I wish to increase the speed of precession tremendously. If I use the formal in (A.) the speed of precession would not be increased, I think. If I keep the 20 X force increase, but lower the speed of wheel rotation, the angle of decent changes toward 180 degrees. I’m not even sure that such an unacceptable angle as that would actually speed up precision.
How can I speed up precession? Give me some ideas and help fellows. |
Date: |
3 February 2010
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Answers (Ordered by Date)
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Answer: |
Harry K. - 03/02/2010 22:17:07
| | Hello Glenn,
The formula to calculate angular velocity of precession is defined by:
wp = T / Lg
(angular velocity precession = tilting torque / angular momentum of rotated mass)
Angular momentum of rotated mass is defined by:
Lg = Jg * wg
(angular momentum of rotated mass = mass inertia of rotated mass * angular velocity of rotated mass)
Some time ago I have sent to you my gyro calculation document with all related equations to calculate angular momentum, precession speed, etc. If you don't find it, I can send it again.
Have again a look to the precession calculation equation:
wp = T / Lg <=> wp = T / Jg * wg
Precession velocity wp will increase, if
1. the tilting torque T would be increased
2. mass inertia Jg would be reduced
3. angular velocity of spinning mass would be reduced
However, there is a limit. You can only increase precession velocity by reducing the anguilar velocity of the spinning mass to a fixed point. This point will be reached under this condition:
angular momentum of spinning mass (Lg) = angular momentum of precessing mass (Lp)
Beyond this point the precessing mass will change to spinning mass and vice versa. That means that beyond this point precession velocity outbalances relating to the spinning velocity and thus the both parameters will be interchanged.
Plaese note that mass inertia of a spinning disc is twice relating to mass inertia of precession velocity!
So you can imagine that the angle of decent cannot change toward 180 degrees, assumed I did understand you correct in this matter.
I hope I could help a little bit. If not, please let me know.
Best regards,
Harry
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Answer: |
Glenn Hawkins - 04/02/2010 00:22:29
| | Dear Harry,
Thank you. I understand your mechanical conclusions. I concur with you for they are also my own findings. Your equations I understand less, but I think they are very special and I can use them with some effort.
I want to take a moment to think you for offering your time and ability. I think people generally would not understand these limitations. In a general sense they are not logical. They become even more complicated in Ravi’s and Kidd’s machines, because directional forces are confined and therefore permit increases in precession speed while still maintaining a 90 degrees angle that doesn’t seem otherwise possible. You are wise to see. Your work in this is very fine.
There is one way to easily increase precession speed, but it is limited also. The longer the shafts the weaker the vertical support during precession, but pointedly the greater precession speed is possible. . but again only up to a point.
Still I must try to find a way to increase precession speed very, very much. That may be impossible.
We’ve had the most snow where I live in twenty years. It made me remember my youth, Spaziergang durch die zwei Meter breit geschwungenen Pfad, der Bürgersteige von Hanau aus hatte geschaufelt. Die Ufer der Schnee auf jeder Seite Hüfte hoch. . schön!
Regards,
Glenn
I may get back to you as I go along. : - )
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Answer: |
Glenn Hawkins - 04/02/2010 23:16:57
| | Hi Harry,
Harry says: “So you can imagine that the angle of decent cannot change toward 180 degrees, assumed I did understand you correct in this matter.”
That was a bad analogy on my part. I should have said, ‘moves in graduated degrees from 1 o toward 180 o, but never reaching 180 o. My poor analogy depicts an arrow moving in a straight line from 0 o to 180 o as you indicate. Thinks for letting me see my error.
Harry: “Please note that mass inertia of a spinning disc is twice relating to mass inertia of precession velocity!”
Would you be so kind as to explain this again?
Here you hit the nail on the head.
Harry: “Precession velocity wp will increase, if
1. the tilting torque T would be increased
2. mass inertia Jg would be reduced
3. angular velocity of spinning mass would be reduced
However, there is a limit.”
I have seen this limit reached in tests using the various criteria of your above statements, (1.), (2.) and (3.). To limit resistance I used crushed ice impacted into the base of the hub and re-frozen. It contacts with a smooth plastic work bench, on which it sits. (The professor’s test was so cheating and he knew it as you argued years ago.) At various ratios of 1.,2.,2., the hub stops following the gyro wheel as in circled in centrifugal response, and begins a binary rotation with the wheel and hub finding a new center axis on the shaft mid way between the hub and wheel. That mid way point becomes the axis of rotation. Your equation is confirmed by these tests.
This problem is beginning to get on my nerves. My effort is like trying to find a way to force square plugs in round holes. You’d think I’d have enough sense to except “However, there is a limit.” -- severe as it is as it is to my design!
Tilting force and angular velocity have a balance point. To reach the optimal precession speed, whenever one of these forces is increased, its paired force must also be increased otherwise a problem with equal and opposite arises. Therefore low paired forces seem to produce the same optimal precession speeds and as when using tremendously increased paired forces. Blaaaaaa! What a bad taste.
Among other things, I still don’t know the key, that is paired calibrations to put into my design. The question is, using your rule:
“angular momentum of spinning mass (Lg) = angular momentum of precessing mass (Lp)
My ‘degree of decent’ doesn’t mean the same as your ‘tilting torque’. The degree of decent is caused by the tilting torque over powering angular momentum. I wish to increase the rate of decent to the optimal limit before opposite rotation of the hub, or pedestal begins. I am searching for a ratio between tilting force and angular momentum before the law of equal and opposite begins to influence the hub.
One more complication to mention is that when I will apply tilting force (consider as downward). I will not add weight to sever as a gravity force to accelerate precession as is usual when playing with our toys. When we add mass we also add inertial resistance in the direction of precession. The mass resistance always exist in all forms of rotation, because acceleration is constant. In normal rotation there is a force to counter this resistance and keep the wheel spinning, but there isn’t a countering force in precession and coasting will not continue once a velocity is reached. Always there will be constant resistance. By adding force without adding mass, greater precession speeds must be possible than our little gravity fed testing indicates, because we are not also adding inertial resistance. We would encounter less limitations to precession speed by applying mechanical force, rather than adding weight, which again, causes hub/pedestal reaction problems at lower speeds than are otherwise possible even at higher speeds. I’m sure, but I’ve never tested it.
Your response separates the men from the boys, but perhaps others with know-how will yet try to help in finalizing my design decisions in order to build.
With appreciation and regards,
Glenn
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
After a nap.
Wow! I just found it. Just now. It’s simple. When tilting torque and angular momentum are equal, the speed of precession will be equal to the speed of tilting decent! Finally as simple as that is I couldn’t find it, until I keep thinking about your post. I can have tremendous precession acceleration while descending at a 45 degrees angle without any equal and opposite interference. I still have a heck of a lot of calibrating to do, but what seemed impossible was not. Glad I saved this for a while before posting.
This is not true after all: “Therefore low paired forces seem to produce the same optimal precession speeds and as when using tremendously increased paired forces. Blaaaaaa! What a bad taste.” We can reach tremendous precession speed without a rearward reaction. We should be happy this evening in solving this problem.
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Answer: |
Glenn Hawkins - 05/02/2010 01:26:25
| | Last communication for a while, I guess you hope. I'll try but I'm having fun again with this for the first time in ages.
I have it all now, Harry. I am really happy, Harry. I’ve stopped just mentioning this design every year, to actually finishing the perpetration to build. I wonder if it will work. I give it better than a 50% possibility. I hope to finish in under 120 days. Take care.
Oh wait. The solution is once again so simple. The degrees of the several disks descending, and the degrees of precession in partial rotation back and forth will be built into the machine and unchangeable, but the solution is to have variable speed motors for disk rotation and separate veritable speed motors to produce tilting torque. You simply adjusts motor speeds and torques to perform to the optimal precession speed that yet avoids a rearward reaction. So simple isn’t it?
Have a nice day tomorrow and please tell Mrs. Harry hello from the states for me.
Glenn
I guess I’m done now, but I will probably run across more problems that you might help with. I don’t know yet.
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Answer: |
Harry K. - 05/02/2010 11:40:07
| | Hello Glenn,
Give me some time. I will answer your postings at this week end.
Harry
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Answer: |
Glenn Hawkins - 05/02/2010 18:02:18
| | Harry, you’ve already helped, both by giving me information and by allowing me someone to talk to while I think out problems. Please forget my posts up to now. I am now progressed beyond those problems and I value your input and help so I don’t want you to waste your abilities on those posts, for as I said I I have already progressed beyond them- - at least temporarily. I do want your help though, very much. I have decisions to make and mechanical methods yet to design, math and ratios to be put down, and there are a lot of selections to be made. Your help could be invaluable. I must rest for a time now. I’m on a crash diet and my energy is shot to hell.
In a few months I would like to partners up with you at %50 and 50% ownership and build this very complicated machine. I can handle the financing and assembly. You would handle the engineering and help refine the design, give new ideas and discuss changes. Maybe you could help a little with financing if you wish and when you know more. There is no demand. Before you consider this I must work more and then arrange to send you pictures and sketches and of course the ideas of it all.
I had a different idea with Ravi. He would have been the foreign purchasing agent for inexpensive parts, labor and the assembly of toys. But my prototype thrust magnesium requires an engineer, not a purchaser.
It will be a good while longer. Hang in. I’m going to find a way to make you my friend again.
Glenn,
You have curiosity, yes? Well I know you do. I go a head and explain one of the theoretical inclusions. Do you believe a 5 LB. rotating disk, precessing at 15 miles an hour is carrying momentum and can deliver its load into a collision, especially under the best circumstances? Note: In the Christmas Lecture, ‘Lecture #15’ do you see there is clearly plenty of momentum shown in the bang and reaction movement upon collision, though in earlier lectures the professor said there was there was little or no momentum and at another time he repeated, “. . . no momentum”? I think you already agree there is. If so, you will be more interested in the plans I have to produce a jerky, but constant re-acceleration into the device, which is based on inelastic collisions. It is as if, the disks kept colliding into an air bags, wherein rearward reaction would be absorbed (no bounce back) and the full load of impact delivered one directionally forward.
Though I believed in my design (not well refined then or now as yet) and I would have tried to built, but for all this flagrant gross misinformation about gyroscopes, so totally wrong. That is why I did not build 20 years ago.
What do you think of all that?
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Answer: |
Harry K. - 06/02/2010 16:53:18
| | Hello Glenn,
Thank you for offering a partnership but unfortunately I'm too much involved with my job since I have been promoted at the beginning of this year. However, I will try to help if you have any technical related questions. I wish you much success with your idea!
You wrote:
"Note: In the Christmas Lecture, ‘Lecture #15’ do you see there is clearly plenty of momentum shown in the bang and reaction movement upon collision, though in earlier lectures the professor said there was there was little or no momentum and at another time he repeated, “. . . no momentum”? I think you already agree there is. "
There is indeed momentum in precession motion and it can be easy calculated:
Lp = JG * wp
JG = mG * r * r
In Christmas Lecture, ‘Lecture #15’ the professor provided the following data:
1. Gyro mass is 8 pounds = 3,63 kg
2. "about 1 revolutions per second at precession" -> wp = 2 * pi * n = 2 * 3,14 * 1 1/s
wp = 6,28 1/s
3. The radius must be estimated and is about 150 mm = 0,15 m
Now the angular momentum can be calculated:
JG = 3,63 kg * 0,15 m * 0,15 m = 0,08167 kg*m*m
Lp = 0,08167 kg*m*m * 6,28 1/s = 0,51 kg*m*m/s = 0,51 Nm/s
The stored kinetic energy can be calculated by the formula:
Ekin = JG / 2 * wG * wG
Ekin = 0,08167 kg*m*m / 2 * 6,28 1/s * 6,28 1/s = 1,61 Nm / J / Ws
That is not "plenty" but also not negligible.
However you can see, the higher the precession velocity, the more kinetic energy is necessary to accelerate the gyro from zero to precession speed. During this acceleration time, the axis of this overhung gyro drops down until precession speed has been reached. Thus the axis drops down faster and faster if precession speed increases due to the decrease of rotation speed of the gyro's spinning mass!
I wish you a nice weekend!
Harry
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Answer: |
Glenn Hawkins - 06/02/2010 21:44:09
| | Thank you, Harry and a nice weekend to you too. Congratulations on your promotion. I'm happy for you, Glenn,
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Answer: |
Glenn Hawkins - 09/02/2010 05:21:45
| | I print this as I continue work on the design, because it may be interesting here.
While I searched for a way to increase precession speed in the design beyond a certain balance point, I recalled forgotten tests I had repeatedly performed long ago.
(((((( I have hit the outer shaft of a spinning overhung gyro on a pedestal really hard with a hammer many times to make sure that sometimes the hammer face had to be near perfectly horizontal. It was a great deal of tilting force. The gyro always shot forward like a bullet, but also the pedestal always flew in the opposite direction. This ‘pedestal flying in the opposite direction’ did not concern me, because the gyro, according to the professors measurements on a sensitive scales weighed hundreds of times than the pedestal weighs. So you see, there was only a tiny amount of rearward inertial resistance furnished by the pedestal, yet the much heavier gyro shot forward. Even if the pedestal had been forced down on the table that wouldn’t account for much frictional resistance as I tested for that too. The conclusion to all this was reached over twenty years ago, but I had forgotten. In summation: The tremendous precession speeds I need seems entirely possible, just as was theorized a few days ago and posted here.)))))
Update: I spent the last four, or five days working off and on, off and on, to find a solution for a very complicated alternating three definitional action. Nothing like it was available and no solutions were available, not even a number of different Geneva gears I inspected came close. I wouldn’t give up. Yesterday I drew it, one of many drawings -- the solution-- is these three gears, an interior driver with positions of certain missing teeth and a plate to hold the two inside planetary gears from moving at certain times. It is complicated in application. Lots of interesting little things are yet to be finished reasoning and finalize. This three-gear solution has taken me years actually, though I only begin earnestly working on it maybe five days ago. I think I must be either one dumb motherfuuuker, or I’ve had a brilliant inspiration. I feel ok. So whichever it was, it didn’t hurt me.
You might like to see an ingenious gear. Notice not only how it changes full rotation into half rotation, but also how it holds the driven gear from moving at intervals. My gear holds that way, but is otherwise very different.
http://en.wikipedia.org/wiki/Geneva_drive
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Answer: |
Glenn Hawkins - 09/02/2010 05:34:35
| | HOLD UP ON ACCEPTING THIS HAMMER TEST. I AM ONLY NOW RECALLING PROBLEMS. Its been so long. I MUST DO IT AGAIN. Sorry.
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Answer: |
Glenn Hawkins - 16/02/2010 17:33:44
| | The biggest surprise I’ve had is from some recent tests that indicate the following.
Generally, if tilting torque is in creased by (1.) while angular momentum is increase by (½) then precession speed increases without causing the pivot area, i.e. string, pedestal area, from experiencing a rearward reaction to precession. (2:1). (I used a drimel tool, ice, timing, weights and various set-ups.) If (2:1) is true I can, with longer shafts and greater force create the tremendous accelerating precession speed my designs require to cause enough thrust to be of real use.
Here are some interesting mechanisms for you to look at.
http://www.flying-pig.co.uk/mechanisms/pages/gearreciprocate.html
Back to the design. I had to stop awhile.
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Answer: |
Glenn Hawkins - 16/02/2010 23:31:38
| | Another interesting gear. I designed this gear for my needs and was happy with myself, because I did not know it had been done before. The driver gear with the handle turns the smaller gear. Notice the missing teeth on the driver gear, then notice the smooth little convex projection on the smaller wheel. This is how the smaller gear is rotated ¾ turn then stopped and held tight for 1/3 turn. These beats the heck out of a stop and go servo motor, which weighs far more and has les power than the continuous drive dc brushless motor.
http://www.unionmillwright.com/typesofgears.html See No. 9. Intermittent Spur Gear Pair
I found another gear pair I had drawn and needed. See No. 8. Single Planetary Gear Set
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http://www.flying-pig.co.uk/mechanisms/pages/gearreciprocate.html
I need both, (Rotary motion to Reciprocating motion) and (Rack and pinion)
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I have more methods, parts and connections. The way it all works is really neat and fun to devise. How to do it is like solving a puzzle. A couple of people here might remember what I mentioned five or six years ago. I said that the motion required in my designs was a nightmare of continuous chaotic tumbling every-which-away, but kept under control. Well I’ve about got it solved, that is how to do it. I’ll let you know how it goes from time to time.
////////////////////////////////////////////////////
Be sure and find a little time to have a little fun through the week. We all read that someone said two thousand years ago, “Live for today for the future is promised to no one.” I think he might have been right, don’t you?
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Answer: |
Glenn Hawkins - 05/03/2010 14:59:49
| | I think I have found all the fundamental solutions to this Rubik's Cube of mechanical problems necessary to make the machine function. At various stages, solutions once found, created the need for more solutions, either that or the abandonment of the solution and the search for another. At stages it seemed that a total combination of workable solutions were perhaps imposable. I think this will be among the most original machines ever built., rivaling, or succeeding perhaps even the first sea going clock. I hope it works.
Glenn,
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Answer: |
Glenn Hawkins - 13/03/2010 17:23:13
| | One thing favoring this design is that it’s very complicated. Any few who may have reasoned out the fundamentals of how it should work, may not have been able to build it or even complete the design of it. This would explain why it hasn’t already been done.
My mind may be too old. I can only concentrate on certain problems for a few minutes at a time. On those problems I have to give up, do other things for hours, go back, and try again and so on and so on. What drives me away is the inability to visualize so many three dimension motions. Steven Hawkings, the world renown physic, indicated he could do that--- sometimes. Well, I think so can I -- sometimes, but not often and certainly not endlessly as the design seems to be driving me to do.
I tire of all the dead end efforts. Over time the near unsolvable problems (for me at least) wear you down with endless failures for every tiny reward of finding a solution. Most solutions are only paths to more dead ends. You cannot force a gyroscopic to do what you want. You are its slave. You must do only what it permits, in only the ways it permits, or it will refuse to work for you. My progress is very slow. For two days I have been drawing six weird looking pieces of metal to allow for a three-combination scissor action. I have the solution, but I don’t have a solid grasp on it. The more difficult my involvement is, the more I believe in the possibility of inertial propulsion. I am however, I should add, making progress even as I complain.
I will fill you in about once a week.
Glenn,
I leave you with this thought.
We may know the laws of motion are true, but I intend to manipulate those laws in confined and forceful ways that would be imposable in nature. Nature would correct my manipulations immediately in the freedom of nature, if those actions ever occurred in nature, but I deny nature a way to unravel the things I do. Inertial propulsion will be a stand alone, contained, artificial condition and the universe and its laws outside this two cubic foot area of a machine will be forever true and unaffected.
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Answer: |
Glenn Hawkins - 21/03/2010 19:36:16
| | As it turned out more design work was necessary.
I believe I have just now finished all the solutions and I am satisfied with them. I think no other solutions would be better.
Now to get all the information about available parts, select and draw it all out with shapes and dimensions for parts that have to be made.
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Answer: |
Glenn Hawkins - 22/03/2010 21:12:14
| | It won’t work. I’m sad. Back to work, sleep, hunting for a solution that will withstand morning after scrutiny.
There are all the reasons I could not complete this design in twenty plus years. I feel like I have all but one solved in this last attempt. This is the closest I have ever come. Back to the hunt.
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Answer: |
Glenn Hawkins - 02/04/2010 22:19:11
| | I've had all the solutions now, all of them, for a couple of days. Something Sandy, finally, finally told me helped me finish the last solution. I have it all. He may too at this point. We will all have to wait and see, while I have to find and buy a lot of stuff now that I know bla, bla and so on, the biggest idem is a milling machine. I have to first patch the holes in my shed's tin roof. Anybody want to buy some rusted out tools and machenery? I've been sick a long, long time, long enough for my roof to rust out- - - and now I am completely well and for many indeterminate years yet to come, God willing.
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Answer: |
patrick - 06/05/2010 01:06:12
| | keep at it or ring me,its worth one call surely
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