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22 August 2019 04:28

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Question

Asked by: Blaze
Subject: Does a gyro wheel retain velocity when released?
Question: My theory is that when an overhung gyro wheel is precessing and is suddenly disconnected from the arm, it will continue to travel in a straight line. This effect would most easily be seen when the gyro is precessing at HIGH speeds. Before I go and "reinvent the wheel" by trying to test this theory, has anyone tested this experimentally or know of anyone who has?

Blaze
Date: 4 June 2012
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Answers (Ordered by Date)


Answer: Ram Firestone - 04/06/2012 20:30:23
 I haven't tried it however I think you are probably correct. It might be kind of a hard experiment to do properly.

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Answer: Nitro - 04/06/2012 22:36:01
 Dear Blaze,

Oh dear! Here I go again showing that I never read the book; “How to win friends and influence people”!

You will not like this but (IMHO) your theory is correct. Bet you weren't expecting that!

Don’t get too excited, though. As the application of a force to alter the axial angle of a (perfect) gyro instantaneously causes it to precess. The removal of that force by, in the example that you instance, its sudden disconnection from the arm will equally instantaneously result in the removal of its cause (force on the axis) of movement (precession) and therefore, instantaneously, leave no reason for its motion to continue. The mass of its ancillary parts (cage, arm etc.) will, however, cause it to continue to travel in a straight line and (some say) as the gyro is no longer a precessing body its own mass will exhibit the Newtonian propensity of “flying off the handle”.

If your present test of the large overhung gyro is setting out to prove that there is momentum and centrifugal force involved in a precessing gyro or if you are just joining the legions of people who have wasted their time in trying to prove the late Laithwaite wrong (somehow trying to prove him wrong is starting to seem almost as crazy as those trying to prove Newton wrong)(it really doesn’t matter whether neither, either or both are wrong) instead of researching your own path with a truly open mind, I am sad.

I am saddened, too, that you might think that there is no need to carry out a test more sophisticated than to “let it "crash" into my hand”. At least let it simply strike an upright (balsa dowel or what have you) that will break with a known weight – to give a true indication of its impact, or why bother.

Here is an easy (math free – for me) web page that can show the force to expect for a given mass and speed and deceleration distance:-

http://hyperphysics.phy-astr.gsu.edu/hbase/carcr2.html

If you can really observe, then here, yet again, are easy to follow video examples by a university (desperately trying to show Laithwaite wrong, so it must surely be regarded as unbiased) showing the lack of momentum and therefor centrifugal force in a precessing gyro.

http://www2.eng.cam.ac.uk/~hemh/gyroscopes/videoseven.html

Please look carefully at video 1. and 3. I would really like to hear your and Harry and Ram’s opinions as to why the precessing gyro does not topple earlier than the static overhung one.

Kind regards
NM


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Answer: Glenn Hawkins - 04/06/2012 22:41:05
 Hi Blaze,
There are two easy ways.
Precess the gyro on a string near the table and drop the string.
Precess the gyro on your finger tip near the table and jerk your finger down.

To the best of my recollections the gyro lands standing still and doesn't move. Odd huh? You are right about fast precession I think. You can do these tests at low rotational speed to produce fast precession. Then comes the argument and pondering about live weight verses dead weight. I hope this is a little helpful.
Glenn,

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Answer: Glenn Hawkins - 04/06/2012 23:11:27
 Well you can hear mine instead. The gyro while precessing dose not force straight down on the pivot as it does when it is NOT rotating. During rotation the torque to the pivot is curving and would cause the the pedestal to rotate around the gyro if freed from the restraint of the table. Whenever the pedestal would respond if it could, as it were about to tip, the curving effect transfers the torque to the very rear of the pedestal base. Therefore the force point is much more rearward and less imbalanced than meets the eye. Ask me any time and I will tell you there is no bogyman in charge of gyroscopes. It is all mechanical. These ideas of no mass, no momentum, no centrifuge and 'you can't precess no-mass' are so ridicules that no wonder none of us are taken seriously in our quest. Harry and Ran I do not speak for, but I thank they would tell you the same if they weren't more diplomatic than I. You are not competent to ostracize a man like me and you shall not.
Glenn

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Answer: Glenn Hawkins - 05/06/2012 01:01:18
 Here are a couple of related things Blaze:

ONE)
While an overhung gyroscope is rotating at high speed, hit the outside axle knob hard with a hammer. Do this several tines trying to aline the face of the hammer completely horizontally. You should see the gyro fly really fast outward into a wall, while the pedestal moves oppositely very weakly.
TWO)
Set a gyro upright on its rim guard leaning slightly away from you. Use a plastic ruler held as horizontally level as possible and shove it very hard with both hands down on the up tilted axle knob. Sometimes the gyro precess too quickly into the ruler. But sometimes the gyro can roll away so very fast. Considering the force of friction resistance of a point of round metal to a plastic surface; in relation to the thrust force of the gyros, it is like maybe 1: 300, or less.

Again I hope you can make something out of this.
Glenn,

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Answer: Glenn Hawkins - 05/06/2012 01:10:18
 Clarification from above:
I wrote: "During rotation the torque to the pivot is curving and would cause the the pedestal to rotate around the gyro if freed from the restraint of the table."

More precisely: The rotation tendency of the pedestal I speak of is NOT horizontal, but vertical rotation as if the pedestal were trying force its way into and through the table. That is curving downward.
Glenn

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Answer: Glenn Hawkins - 05/06/2012 02:53:48
 1974-75 Christmas Lecturs
7. Air powered gyroscope (5000rpm - 8lb). Searching for centrifugal force. Gyroscope hanging over the top of a table. Out of balance by 2kg.

When the gyroscope on the table is not precessing, it's tipping over force is not at the center of the base, but at the very front edge of the pedestal’s base. This extended point of balance is beyond the false appearance to the eye and judgment.

When the gyroscope is precessing, deflections at the top and bottom twist the gyroscope's weight onto the pedestal. The twisting point of force is at the very rear of the pedestal base, not at the center of the base. This receded point of balance rearward to the center of the base is beyond the false appearance to the eye and judgment.

The same conditions hold true when the gyroscope is upside-down.

The illusion as to the places of the points of balance for both still, and active gyroscopes has fooled everyone. So don't feel bad.

Don't be mad at me, please. I am what I am, which is mostly good, Nitro got me on fire and Ram exasperated me, but not any more.
Glenn,

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Answer: Blaze - 05/06/2012 04:21:34
 Thank you all for your responses gentlemen. I value your input.

Gentlemen, there is only one person that I am trying to prove right or wrong. That person is me and no one else. I have certain theories based on science, math, the little I do know about gyros unusual characteristics and a tremendous amount of thought about what I think should happen with gyros. Now I am trying to see if my theories are correct or not.

Ram, I agree it would be difficult to do this experiment well.

Nitro, all I was really trying to prove was that the mass is still there in a precessing gyro, however, I was also thinking that getting a better idea on a number on stopping force with the giant gyro might be beneficial. It is easy enough to spin it up to speed now so further tests shouldn’t be too difficult. I thought of a “Newton’s Cradle” approach where I would hang a heavy weight by a string and let the gyro crash into it, although that may only give me a better idea rather than a hard number. Your idea of breaking a dowel might also work. In either case the gyro would lose all or at least a significant portion of its speed and drop, so I would have to be prepared for that. The problem with breaking a dowel may in its repeatability. The force to break it may not be consistent and may depend on which way the grain of the wood was in relation to the force hitting it.

The precessing gyro has to be further out to topple in video one than the non precessing gyro in video three simply because the WEIGHT (not mass) of the precessing gyro is transferred to the pivot, so in the precessing gyro it is as if the weight of the gyro were sitting on top of the pivot. When the gyro wheel is locked overhanging the table, the center of mass has also been shifted further out past the table edge. The gyro base is bolted to the plate and the non precessing gyro locked in place caused the shift in center of mass..

Glenn, I too have thought about how the force down on the pivot would cause movement if the pivot could curve downwards. I believe that if the gyro was precessing while floating in space (an artificial down force would have to be supplied) on a small platform with the pivot bolted to the platform, the platform would “corkscrew”. A mirror image system or systems would have to be used to prevent this or other unwanted corkscrewing effects. I will try the experiments you suggested and a few of my own too.

So, I take it that no one knows if anyone has set up an experiment where the gyro wheel was suddenly released from the arm or pivot with the express purpose of seeing if the gyro wheel continues on in a linear fashion?

Blaze


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Answer: Glenn Hawkins - 05/06/2012 05:14:02
 I did. I explained.



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Answer: Blaze - 05/06/2012 19:12:20
 Thanks Glenn. Your experiments give me some of the answers I am looking for and I will try to repeat your experiments to make sure I understand the results properly. I am looking for something a little more specific for what I have in mind though, and that would take a gyro wheel moving at several feet per second to be released, so I will try to set that up for an experiment.

Blaze

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Answer: Sandy Kidd - 05/06/2012 23:24:55
 Glenn,
You said:.
“Ask me any time and I will tell you there is no bogyman in charge of gyroscopes. It is all mechanical. These ideas of no mass, no momentum, no centrifuge and 'you can't precess no-mass' are so ridicules that no wonder none of us are taken
seriously in our quest.”

All of these claims were made by me Glenn, as you very well know.
There is no bogeyman, they are purely mechanical, but whoever said otherwise?
Why should we be taken seriously?
You are quite happy to accept inertial thrust only if it is done respectably.
If you are under the illusion that any of this stuff is taken seriously anyway you are deluding yourself..
The point is Glenn, you choose a course, you keep to the course, you stay the course and if things turn out to be a bit unorthodox you have to accept the inevitable.
I never made the rules.
I said that gyroscopic action is a state of mind and I mean this more today than ever.

You admit that a disc in precession demonstrates no angular momentum and or centrifugal force, because it is balanced out by the action of the gyroscope.
If you are really in doubt about this it is easily proved by measuring the disappearance of centrifugal force.
As you know this is relatively easy to confirm.
There can be no angular momentum if there is no centrifugal force.
You may have to do this first, before re-joining the discussion.
After you have proved there is no angular momentum, and the thing is still rotating what has happened to the mass?
With mechanically accelerated systems all this stuff is much more pronounced than in a relatively slowly rotating system in precession, for as part of something which can be rotating quite rapidly, this mass does not appear to be doing a lot does it?
Where is the acceleration Glenn, any kind of acceleration Glenn?
Inwards and upwards flywheel acceleration excepted.
Blaze and Luis will testify mass is always there and we only transfer weight.
So would I have twenty or so years ago.
Their way is probably the classical way but this is gyroscopes and things are a mite different.
Maybe it is not respectable Glenn, but unfortunately this is all about spinning discs and prior understanding of these things was zero up until we got involved
Harry and Ram’s opinion will probably be the same as yours, and may be interesting but whatever it is, will not change anything.

Ridiculous you say?
Twenty or so, years ago.my first and successful machine was borrowed from me in good faith.
In reality, that meant being borrowed by a bunch of lying illegitimates.
My device appeared without me on “Tomorrow’s World” where I was duly ridiculed in my absence in front of many millions.
Sandy Kidd.


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Answer: Nitro - 06/06/2012 01:03:13
 Dear Blaze (and Glen – in case he feels left out),

>>Nitro, all I was really trying to prove was that the mass is still there in a precessing gyro

I am sorry, I thought I had made clear in earlier posts that it is the “effects” of mass (and for the pedants among us, should there be any, that should probably read “most of the effects...”) that disappear. Short of having an education at “Hogwarts” (and, yes, mine was close with A S Neil making a very good double for Dumbledore) there is no way that the actual mass will disappear (though that Higgs is still bloody elusive ). I also said, perhaps not clearly enough, that I believe the impact of your test will be likely to equal the weight of the stationary gyro. That is plus,of course, the inertial force caused by the deceleration the non spinning mass (i.e. I do not expect you to find any inertial force from the deceleration of the gyro’s mass added to that of its weight) of its arm, support, cage etc., the instant the gyro’s end (the bit you are catching) meets a fully immovable object. At that instant, if it is also prevented from dropping, it ceases to behave as a gyro (because it is no longer free to precess) and transfers its weight (and the arm etc’s inertia) to the immovable object. If it is free to drop, it then effectively precesses this impact (its own weight and the arm etc’s inertia.) through ninety degrees and, as you know, drops like a stone or, to be more precise, a non precessing wheel.

Naturally, like most visiting this site, I have preconceive ideas on this (and almost everything else in the universe – most probably wrong, I know!) but I will be glad to have any of my misconceptions dismissed by results of good experimentation if this helps better understanding.

The very point that I was trying to make with the university video link I last put up was to illustrate that the expected momentum, inertia and its concomitant centrifugal force from the rotation of the “still there” mass of the gyro is “EFFECTIVELY” missing, precisely because this, along with its weight caused by gravity, acting on its mass is precessed i.e. transferred to the support/pivot that is attempting to change its axial angle with the help of gravity. Kindly note the “attempting”.*

For many of us, nothing but the key is missing. It is my belief that this missing “effect” is the key to our search. That should, of course, read “my search” as you may all be looking for something else – my search tonight will largely be for a nice old fashioned pub with soft leather arm chairs and Adnams Ales so that I can toast Her Majesty's Jubilee in great comfort. It is nice, though probably a bit nasty as well, to know that you, so lucky in most things, Americans just occasionally miss out on a few of our best things. I hope that at least some of you drew enjoyment from our celebrations on TV.

To see that there really is something anomalous (or, if you prefer; “another crap unprovable test”) see the video that I put up earlier that you may not have seen – it may at least give you a good laugh!?

http://www.youtube.com/watch?v=xfyJbAP1cR0

Glen
>>You are not competent to ostracize a man like me and you shall not.......


How competent, and in what, does one have to be to ostracise someone, I wonder? Anyway, you daft and confused old American you, I am not trying to ostracise you (though the comments you make sometimes are very off-putting and seem to seek that very end, like my youngest child did when pushing boundaries and seeking assurance at the same time that we still loved him) and anyway, if I were Royal Arch, I couldn’t.

Kind regards,
Love and laughter
NM
* just reread this on return from the pub (never a good time to reread anything) and have decided to award a medal to anyone who can wade through this with full understanding – don’t think I am going to get a medal, somehow!

PS (ed as a newt I hear you cry) Hi Sandy. Say hello to my ancestral home of Mull for me.



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Answer: Glenn Hawkins - 06/06/2012 02:23:13
 Dear Blaze.
I have a lot more suggestions that you don't need and here is a list . . . :-)
I think I will immerse myself into several things now and just occasionally watch what you all are doing and saying. Wishing you very good luck and bye for a while,
Glenn,

Dear Sandy
I am quick and hard headed and sometimes need to beaten accordingly. I am hurt by what I said. I hurt. You are to be admired, not spoken against and my stupid word 'Ridiculous' you are not, but I am for using it. For seven years I have felt differently than you on 'SOME' things, but I was never I less than appreciating and nice. That was because of my respect for you and the huge body of work you have done. That and the fact you are just simply easy to like and I like you. If I could let you sock me in the nose, I would. I am sorry.
Glenn,

Hello you sweet old sot, you old video maker you,
What were you celebrating? Just kidding. By the way you owe me almost a half of a medal. A few brews does improve one's perspective self examination, hay? Nice self critique. Actually I easily understand you despite the accent. I could really use a little love today, Nitro. How kind of you. What a day! What a day I have had. Peace and good will,
Glenn,

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Answer: Blaze - 06/06/2012 04:14:29
 Hi Nitro. Thanks for all your input.

“As the application of a force to alter the axial angle of a (perfect) gyro instantaneously causes it to precess. The removal of that force by, in the example that you instance, its sudden disconnection from the arm will equally instantaneously result in the removal of its cause (force on the axis) of movement (precession) and therefore, instantaneously, leave no reason for its motion to continue.”

Normally that would be true, but this scenario is not the same thing as a normally precessing gyro. In a precessing gyro the movement around the pivot stops when the tilting force is removed because the momentum of the precession is used up regaining potential energy. Theoretically during the Stop you should regain all the potential energy that was turned into kinetic energy during the Drop.

When precessing, the gyro wheel always keeps one face directed towards the pivot. So for 360 degrees of precession, the gyro also rotates around a vertical axis through its center. If the gyro were suddenly disconnected from the pivot arm, the gyro would still stop its rotation through its vertical axis because the tilting force has been removed. However, the precession speed would not be able to act against anything. There would be no regaining of potential energy, no rising of the gyro wheel, because neither axle end would have anything to act against to generate the rising of the gyro wheel.

The speeds of table top gyros are usually less than a few inches per second. To see any continued movement before it falls down and hits the table would be difficult. I want 4+ feet per second movement during my experiment with a starting height of about 6 feet. That should give me a forward movement of a foot or two before the gyro hits the ground. That should be enough movement to actually see, well, at least that is the theory anyway.

So I agree that precession would stop, where precession is the rotation around a vertical axis through the center of the gyro wheel. I disagree that all forward motion of the gyro wheel would stop when suddenly disconnected from the gyro arm, but that is what I want to test. We shall see.

Blaze.


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Answer: Ram Firestone - 06/06/2012 19:15:58
 To clarify my statement a bit, as I said I do believe the gyroscope would continue in a straight line however (and I’m sure I’ll get verbally beaten for this next part) this is not a reactionless process since if the stand were on a frictionless surface it would be pushed in the oppose direction. If the stand had enough friction that it wasn’t moving in the first place, the gyroscope would essentially be pushing off the earth.

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Answer: Glenn Hawkins - 06/06/2012 22:20:47
 Dear Ram,
Send me an email with your post office address and I will send you a big box of wonderful chocolates. I am very serious, because I am sorry.
ehawkins32@comcast.net
Glenn,

Blaze my friend, you are reinventing logic and physics, but I know you must. It is interesting still and at this point even neccisary.
Glenn

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Answer: Sandy Kidd - 06/06/2012 22:56:46
 Nitro,
I was interested in your comment relating to your ancestral home on the Scottish West Coast.
In light of that I have been unsuccessfully trying to trace the MacMad clan in your ancestral home of Mull.
Even had a word with Paul McCartney over a couple of pints in a local hostelry, but he says he doesn’t know of any MacMads in the Kintyre area
I searched for Minis with wrecked Cooper “S” engines, and burned out tyres with no luck..
My last option was to source purveyors of large quantities of nitro methane who would surely point me to the MacMads.
They must have been a rich bunch the MacMads to afford to run sprint Minis on that expensive stuff.
I had previously considered that Sir Paul was the only person in Kintyre who could afford to feed this stuff to a guzzling “S” through a pair of 45 twin choke Webers.
Anyway I did not find any so I gave up.
Rumour has it that they got PS ed off with all that fog rolling in from the sea and moved out.
I think they have all gone to the States, Oz, or some even to Englandshire.
Still hunting through the cemeteries in Kintyre though
At least one of them must have snuffed it at home..
Sandy.
PS Has the prodded gull got any interest in the spinning wheels.


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Answer: Blaze - 07/06/2012 01:03:02
 Hi Ram. Thanks for your input.

“if the stand were on a frictionless surface it would be pushed in the oppose direction”
Certainly if this were a “rock on a string” that was circling around the pivot, that would be true. However (and this is just my theory) I believe that that won’t happen when a gyro wheel precessing at steady state speed is released from the arm because the gyro wheel is just “coasting”. There is very little if any reaction at the pivot during steady state precession, unless the precession period is fast enough that the "slinging" effect of the dead mass starts making itself felt. The reaction at the pivot has already taken place during the Drop when the pivot does experience a horizontal force for the duration of the Drop. Once the gyro is precessing at steady state speed it is coasting with no driving force to make it coast, thus the term coasting. The precessing gyro has stored momentum that can be released anywhere from 0 to 360 degrees in the precession plain. Because the wheel is coasting, the pivot won’t experience a reaction opposite to the direction the wheel travels when the wheel is released, like it would if a rock were being released. Well, at least that is my theory. We will see if I am right or wrong.

Blaze


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Answer: Luis Gonzalez - 07/06/2012 02:47:51
 Hi Blaze,
Once again, I am in agreement with you Blaze.
Precession has its inner “rotation-like” motion, and its external “orbiting-like” motion.
The flywheel’s “rotation-like” motion stops instantaneously when the tilting torque is removed.
However the external “orbiting-like” motion has some degree of momentum.
I hope I have paraphrased the meaning accurately.

I also hope your experiment will help to determine how much momentum of the latter motion has.

Best Regards,
Luis G

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Answer: Blaze - 07/06/2012 04:08:10
 So I came up with some preliminary parameters for my experiment which I hope to perform in the next week or so, if all goes well. I will be using a 3 foot or 4 foot arm, somewhere around a 12 to 15 inch diameter wheel, a precession period of just over 2 seconds which should give me a speed at the wheel of 9 to 12 feet per second. When starting from a 6 foot height this should give me 5 to 7 feet of linear movement before it crashes into the ground, well, at least that is the theory anyway. I will refine these parameters over the next few days before building anything. This may be a destructive test so I want to try to get it right the first time.

Cheers,
Blaze

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Answer: Nitro - 07/06/2012 21:42:55
 Dear Sandy,
Ha
Ha
Har
Har
Hahaha
Cough!
Splutter!
HaHahaha.
Thud!


Now look what you’ve made me do! I’ve fallen off my chair and broken my coccyx (that really smarts!) my arms and two teeth. Luckily, having spent so much time with gyros, I have evolved gimbal hands and I haven’t spilled a drop of my Glenmorangie, so I am alright! Notice I said “hands”! Weyhey! Stereo, gimballed, scotch!

You must have done a load of research to come up with that masterpiece, Sandy, but then, as they say in the L’Oreal adverts “I’m worth it”. Just goes to show that too much use of L’Oreal products risk making your head swell up, I think (that’s to prevent being sued), if you’re not careful! In truth, Sandy, my clan hail not from the Mull (for you of the US of A; a Mull is a corridor of water between the two bits of land – usually between the mainland of Scotland and an island) of Kintyre but, instead, from the Island of Mull (I guess the Isle of Mull’s Mull is called the Mull of Mull!?)(I think I’ll just Mull that over for a while, ‘till my head stops hurting!). There are bones of the ancient MacMads in the graveyard near Craignure, on the road to Duart Castle and there is a cairn near there raised to the memory of my great great uncle Hugh who was a reputed to be a better poet and lover than that Rabbi (intentional) Burns character (Oy Vey! What you mean; he’s not Jewish!? What’s that “A man’s a man faw au tha’” poem all about, if it’s not about circumcision?).

Unfortunately my poetic ancestor made the mistake of writing in pure Gaelic instead of fake “Scotslish” so was never heard of again. Proof, that you should always follow a trend – or you will be left behind with an extinct language, a Betamax, or eight track, or Ariel Arrow, or even gallons of Nitro Methane (invest in commodities in time of financial turmoil, I always say. – What do you mean? Nitro is not a commodity!) and an ”S”. FYI The “S” had twin 1 1/2” SUs (eat your heart out Carol Shelby – BTW, The bruv, Ag, built Shelby’s GT40’s for Ford in Slough – wish he’d nicked one for me) with 1/8” hand shaped ballistic needles made from old bits of brazing rod - crude has always worked for me. I was, however, helped by my brother, Ag (his engineering is not crude), maker of the astonishing Jade Warrior sprint bike, (u-tube that) who managed to procure for me sticky practice (10” Mini) tyres from the six wheeled F1 Tyrrel. In exchange I gave him my teaching of negative ground effect (like the reverse of Sir Christopher Cockerel’s invention) that is used universally now to improve adhesion without weight. Why do I always patent the wrong ideas?!
The “prodded gull” has found a lovely female nest partner so is a little too preoccupied to think of spinning mass, in fact knowing that young men have not sufficient blood for two items to work simultaneously I am not sure he been able to think of anything much recently – lucky sod!

Kind regards
Love and laughter
NM


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Answer: m - 08/06/2012 12:35:56
 

Blaze,

you said “My theory is that when an overhung gyro wheel is precessing and is suddenly disconnected from the arm, it will continue to travel in a straight line.”

That is the “does it have momentum” question, asked in a different way. Nitro’s answer is a good as you will get. No momentum. Or as us pedants like to say, the dynamic component of a spinning disk does not exhibit linear momentum. I gave details of two experiments which demonstrate this in an earlier thread.

The walking gyroscope, posted previously at length, gives an effect similar to the one you are proposing, but without the fear of death distraction which seems to be a dominant feature of your experimental technique.

As the weight is transferred rapidly from one parallel bar to the other, the arcing mass should cause the supporting cord to pendulum forward. It does not. Nor does it swing backwards as the mass is accelerated into a fresh arc.

Nitro,

Video 7. It does not overbalance because the gyroscope couple transfers the reaction to the base. Like Blaze said.

Laithwaite intended that video 7 should be an example of absent centripetal force. With the apparatus balanced on the edge of the desk, as the mass swings around, the reaction to the centripetal acceleration should pull the top of the tower outwards, toppling the whole thing.

Regarding the Cambridge University video link. The most astonishing thing about it is that it is still there!! If you take the trouble to read the preamble by Emma Wilson, it is clearly meant to demonstrate a heavy gyroscope with a light tower. Simple scaling from the video mandates a tower weight of more than 10 times that of the gyroscope to place the C of G in the position claimed.

I did write to Dr Hugh Hunt, to ask his opinion. As you no doubt expect, no reply, no acknowledgement received.

The missing “effect” is Dark Motion, the linear equivalent of gyroscope precession. It can be derived from first principals with the same method as Gyrodynamics. I am currently attempting (with no result) to interest the scientific community.

Momentus


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Answer: Blaze - 09/06/2012 15:50:01
 Momentus, so you and Nitro are saying that if a wheel (50 pound wheel moving at 10 feet per second for example) is suddenly disconnected from the arm while precessing, while it is free to move through the air in any direction, without anything it act against, and with no forces to push against its direction of travel, it will just stop and drop? Now that is something I have to see to believe.

Blaze

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Answer: Glenn Hawkins - 09/06/2012 19:20:57
 This is 50 lbs traveling at 7 mph, which is a very, very fast walk. You have built a heck of an argument, outstanding! but I don't have to see it to be convinced the wheel will not stop in mid air. Neither do you I feel pretty comfortable in saying, but it is fun following you and there is a good purpose. Some reasoning the test will receive will involve dead weight as the culprit of actual force and movement. But I believe it is all weight, both live and dead that will carry the total amount of released speed and momentum that the simple class room equations tell us.

These are such excellent things you are doing. If you could test and measure, both spinning disk and non-spinning disk with a convincing approach, you would prove a point of long and great discussion among our clan here.

I do not know an easy way to set this up, Blaze. For me, the methods are difficult to build. You tell us how you intend to go about it and we might come up with some good ideas. If we find we can't help you, you lose nothing.

Glenn,

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Answer: Blaze - 09/06/2012 21:26:25
 You are right Glenn, I am convinced that the wheel will not stop in mid air for the reasons I posted earlier in this thread, but I am going to test it anyway. I have been working in the garage today and have narrowed down my parameters a bit more. I will be using a 12 inch diameter bicycle wheel (weight of 3 pounds) with a light weight 24 inch arm made mostly of aluminum tubing (total weight of arm will be about ½ pound). When precessing with a period of just over 2 seconds this gives me just over 6 feet per second and a calculated distance of about 3.6 feet of linear movement before it hits the dirt when released from a height of 6 feet.

Unfortunately I cannot come up with an easy way of disconnecting at the wheel end of the arm (I have some ideas, but none of them are easily set up) so I am going to use a string to “hang” the pivot and let go of the string for the release which will drag the lightweight arm along with the wheel. I realize this will skew the results somewhat but since the arm is considerably lighter than the spinning mass I believe I will still get reasonably valid data.

If the wheel does in fact stop, as some people think, then the dragging the arm along won’t matter as the gyro action would overpower the arm’s continuing movement anyway. If my assertion is correct about continued movement of the wheel, then the arm weight will actually slow the wheel’s continued movement a bit as at least part of the arm near the pivot has to be accelerated to the wheel’s velocity.

I like your idea of a spinning and non-spinning test. I will do both, but the spinning one first because after the landing the wheel may not be in good enough shape to spin again but if it is still in one piece it won’t matter for the non-spinning test.

Cheers,
Blaze


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Answer: Glenn Hawkins - 09/06/2012 22:53:58
 Dear Blaze, I don't think it will work. I think your results will end up supporting Nitro's and Momentous' contentions, instead of what you seek. I have reasons to think this way.

Let us try something else if I may join you in thought. The wheel needs to set at the edge of the shaft. The shaft's tip may have gears, or an octagonal shape. You are going to have to have a spring loaded, or gun powder, or electromechanics reverse polarity readied to release and shove the wheel off the axle at the right time. There are other ways, but the wheel has to be separated from any drag at all. Pleas let me know what you decide.

Cheers Glenn

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Answer: Blaze - 10/06/2012 00:04:13
 Hi Glenn. I agree that the best experiment would be to have the wheel disengage from the arm. Do you think it won’t work because the arm will skew the results to the point of making the experiment totally invalid or for some other reason?

If you think it won’t work because the arm will be dragged along, then consider that if the arm were still connected to the pivot and gravity was “removed”, the gyro would still have to stop the arm’s movement and it would as we both know. So if the arm is attached to the wheel when I let go of the pivot, the wheel would still have to overcome the “non-movement” of the portion of the arm that is near the pivot. The two situations, although opposite of each other, required the wheel to positively accelerate or negatively accelerate the arm the same amount, so I still think it should still work.

I have completed my build. I ended up with a 21.5 inch distance from the center of the wheel to the where I will be attaching the string for the pivot. This should still give me 3 feet of horizontal distance if my theory holds true. Part of my preparation was to take the bearings out of the bicycle wheel and clean them up and oil them with a "superlube" that I have. It sure spins nice now and it takes a long time to slow down. Just so nothing is left to chance, I tested it precessing by mounting it on my giant gyro pivot. That pivot sure comes in handy.

If it weren’t raining cats and dogs here I would have already tested my theory by releasing string at the pivot end and let the whole thing fly. As you can imagine, this has to be an outdoor test. So, sometime in the next few days I will do the test by releasing at the pivot as originally planned unless you can talk me out of it before the test. I do have extra bicycle wheels, so if necessary I can build another one and try to work out some sort of release mechanism and repeat the test but I believe that any release mechanism will be complicated.

Cheers,
Blaze


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Answer: Glenn Hawkins - 10/06/2012 01:11:52
 Hi Blase,
There is a reason I am worried about your test. First, I could be wrong. Allow me to think out-loud. The wheel will be traveling much faster than the short end of the shaft and in graduated incremental through out it's length and the pivot is not moving at all, except pivoting. The shaft will drag against the wheel it seems. This dragging can twist the wheel. When wheels are twisted the plain of angular momentum is forced off course. The wheel resist this in a way I wrongly call deflections. These deflections act like breaks for everyone knows (that's not right) inertia dose not want to change course. Now I know this doesn’t seem entirely sensible thinking, but I have experiences it with a gyro on a stick--so to speak. The gyro touches down and rolls and it is applying it's own breaks as it rolls. Can it roll in air?

OK I've blabbed out loud and won't clean it up this time.

Maybe this is non-sense, but I have experience some strange things just like the rest of the clan are talking about. You have nothing to lose. I think you will win as is, but if the thing acts crazy and doesn’t give you what we both basically believe in, then more and more elaborate testing is in order.

When the rain stops! By-the-way where are you?
Glenn

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Answer: Blaze - 10/06/2012 06:32:47
 Hi Glenn. I have already thought of what you describe. The arm near the end of the pivot will have to be accelerated up to the speed of the wheel which will cause a “gyro action” which will in fact cause the wheel to precess. The arm at the pivot end should in fact rotate down, or to put another way, the wheel will precess through an axis in the horizontal plain. At the same time there will be a slight decrease the forward movement that will happen because the pivot end of the arm has to be accelerated. So I may end up with a two or three axis tumble but the forward speed won’t change significantly. I calculate about 1/2 of a foot per second of horizontal movement less than the release speed of about 5.5 feet per second.

I also hope to get a video or two of this, one from above and maybe one from the side at some safe distance. I will probably mark up a 2x4 with markings every 6 inches and set it on the ground for a reference and/or to scale distance in the video from above.

I am in the province of Saskatchewan, Canada, and I dare you to pronounce the province name correctly. ;-)

Cheers,
Blaze


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Answer: Harry K. - 10/06/2012 17:34:46
 Hello Blaze,

Nitro wrote:
"...At that instant, if it is also prevented from dropping, it ceases to behave as a gyro (because it is no longer free to precess) and transfers its weight (and the arm etc’s inertia) to the immovable object. If it is free to drop, it then effectively precesses this impact (its own weight and the arm etc’s inertia.) through ninety degrees and, as you know, drops like a stone or, to be more precise, a non precessing wheel. ..."

I think similar in this matter. The precessing gyro has stored angular momentum in its precessing movement. If you could cut off the spinning gyro from the end of its pivoting lever arm, the stored angular momentum in precession plane would cause the gyro to "deflect" in 90° to the precession plane, i.e. the spinning gyro would rotate around its centre of mass about 90° to the precessing plane until the stored angular momentum (stored in precession plane) is consumed. At the same time the spinning gyro will fall down in the same way like a non spinning dead weight mass.
The reason is the missing cause for precession movement as soon as the the gyro will be cut off from the pivoting lever arm, thus only the stored angular momentum in precession is available, however it will not be tranferred into linear movement but in deflected rotation around the centre of spinning mass.

Regards,
Harry

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Answer: Blaze - 10/06/2012 18:50:04
 Hi Harry, thanks for your input.

Actually Harry, if I understand you correctly, the gyro won't do what you think. When the gyro is cut off suddenly from the arm it just stops precessing, it doesn’t change its direction of precession (where precession is described as the wheel moving around any axis through its center of mass). You can do this experiment easily with any toy gyroscope. I use the Tedco Precision Gyroscope but I imagine anyone would work. Spin up the gyro, hold the pivot in one hand and place the gyro on the pivot. While the gyro is precessing, slowly raise your arm to its full extent overhead, then very quickly move your arm down so that the pivot is no longer in contact with the gyro. You will see that the gyro appears to “instantly” stop precessing and appears to fall straight down. The reason the gyro stops precessing is because when you pull the pivot down the forces are balanced on both ends of the axle. It doesn’t matter if the balanced forces are both pushing up or both pushing down (gravity during free fall while the toy gyro is falling to the floor). As long as the forces on both ends of the axle are equal, the gyro stops precessing. I have done this tests many times. The problem with determining whether the toy gyro continues in a linear fashion is that during a drop of 6 feet you would only get a forward movement of about 1 or 2 inches, not enough to see easily with the naked eye.

Cheers,
Blaze


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Answer: Ram Firestone - 10/06/2012 21:47:02
 " the spinning gyro would rotate around its centre of mass about 90° to the precessing plane until the stored angular momentum (stored in precession plane) is consumed. At the same time the spinning gyro will fall down in the same way like a non spinning dead weight mass. "

I was actually thinking the same thing after I gave it more thought. However I still think the while gyro will continue on in a tangential course while doing this, as will the base in the opposite direction.

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Answer: Momentus - 11/06/2012 10:41:33
 Hi Blaze,

No linear momentum is stored in the spinning disk, only angular momentum. Perhaps this will be clear to you after you have completed your experiment.

BTW If you lift and release the support cord rather that cut it, that will give you freefall, and bring the wheel to a halt before impact with the ground, you will be able to repeat the experiment. Also, if you video this you will see that the wheel starts to move forward again “instantly”. (after the vertical, no forward momentum, drop).

Single instances of an experiment are seldom reliable, particularly as with a gyroscope the damn thing never does what you expected it to the first time, or the second time for that matter.

Good Luck

Momentus



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Answer: Glenn Hawkins - 11/06/2012 13:20:09
 How fascinating to see such intelligent, well trained men locked in dispute over the action of a wheel invented in 3,500 BC. I have seen and registered what Momentous, Nitro and Sandy Kid refer to. But, Blaze is multiplying speed times mass to equal ten thousand times more momentum than our toys generate.

Rotation is a condition of linear motion denied. Every particle is in a constant state of seeking to break free and travel in a straight line. If you were to whorl a hornets nest inside a big balloon, it would make no difference how the angry bees were flying around inside. In that respect, precession is like a sling shot in motion waiting to be released.

We shall see said the blind man. I believe everything depends on the test method.
Glenn,


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Answer: Ram Firestone - 11/06/2012 16:44:15
 "No linear momentum is stored in the spinning disk, only angular momentum. Perhaps this will be clear to you after you have completed your experiment."

If a spinning disk breaks apart are you saying pieces will not fly off in various directions?

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Answer: Blaze - 11/06/2012 19:51:50
 Hi Momentus. Thank you for your input.

I don't believe I said that linear momentum is stored in the spinning disk but rather that momentum is being stored in the movement of the wheel around the pivot (orbit like motion) and when wheel is released and free to go through the air it will travel in a linear direction.

You bring up an interesting variation to the experiment but I'm not sure that it would give a good result. The pivot would have to angle the string when trying to move off in a linear fashion. If the string is short this would cause the pivot to rise a fair bit and would therefore be working against gravity. In my humble opinion, this would skew the results quite a bit. However, I may try it anyway as it is simple enough to do.

I believe that to see some of the effects we are looking for, a larger version of the overhung gyro is required and that is why I have been building these monsters. The toy gyros are quite small and therefore are too easily overcome by frictional effects of what ever apparatus is being used to see the effect.

We shall see sometime this week what happens. You may be right, I may be right, we shall see.

Blaze

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Answer: Ram Firestone - 11/06/2012 20:25:22
 One more thing to consider is that the initial question stated an "overhung gyro" which typically means one that is already NOT precessing around its center of mass. If a gyroscope was precessing due to other forces besides gravity AND also precessing around the center of mass (of all moving components) I don't think it would go anywhere.

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Answer: Blaze - 11/06/2012 23:27:29
 Hi Ram, thanks for your input.

I agree that this test would not work on a gyro that is NOT an overhung gyro. The gyro would not move in a linear fashion when dropped as there would be no "orbital" motion.

I plan on doing this test tomorrow evening at 7:00 PM my time, weather permitting. I have lined up a crew of 3 additional people that wiil be aiding. I will be spining up the gyro and get it precessing smoothly, two people will be on cameras, one from above, one from the side at a safe distance, and one person will actually be dropping the string attached to the pivot. If the thing is still in one piece afterwards, I will do a non spinning test to see if the results are a similar distance to the spinning test. The starting height will be 6 feet above the ground and the linear travel distance I expect is 3 feet. The person doing the drop and the above view camera man will both be in my treehouse and therefore should be well protected should anything unexpected happen.

wish me luck (I'll probably need it),
Blaze

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Answer: Glenn Hawkins - 12/06/2012 01:12:37
 Hoorah, hoorah, hoorah, Blaze.
Nail your colors colors to the mast, To fight or hold out until the bitter end; to refuse to compromise, concede, or surrender; to persist or remain steadfast, especially in the face of seemingly overwhelming opposition.
Hoorah, hoorah, hoorah, Blaze.

Glenn,

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Answer: Harry K. - 12/06/2012 19:14:24
 Hello Blaze,

I rethought what I wrote and I have to consider that I made a mistake or in better words I disagreed with myself what I always stated before. ;-).
A precessing overhung gyro in orbital motion stores angular momentum caused by its dead weight mass in addition to the stored angular momentum in the movement around its centre of spinning mass.
However, to accelerate the overhung gyro from zero to precession speed, energy is necessary. This energy will be gained by potential energy during the drop of the overhung gyro.
This potential energy caused by the drop is equal to the kinetic energy stored in the precessing movement and thus you will not get more than this amount of energy back.

Let's see what your results will be, I'm very curious.
Good luck and be careful!

Regards,
Harry

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Answer: Blaze - 12/06/2012 19:42:13
 Hi Harry, thanks for your input.

"This potential energy caused by the drop is equal to the kinetic energy stored in the precessing movement..."

Yes that is how I see it too.

"...and thus you will not get more than this amount of energy back."

Correct, however, I never expected to get more back. You can't get something for nothing.

Time to the experiment is T minus 6.5 hours and counting, providing the weather holds (they are predicting afternoon and evening thundershowers here).

cheers,
Blaze



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Answer: Ram Firestone - 12/06/2012 20:39:08
 Energy is only part of the story however. Even if everything obeys conservation of energy that still doesn’t make something reactionless. To me any test that pushes off something that is in any way fixed to the earth does not prove something is reactionless. That includes hanging strings which are partially fixed by gravity, or stands fixed by friction. Tests done with minimal connection to the earth are in my view the only good tests to ascertain if a process is reactionless. As I said many times before, nobody in the scientific community will accept any successful test of a non-Newtonian drive when they can in two seconds point their finger to a simple Newtonian explanation for the results. Why should they? That being said tests that are helpful in understanding gyroscopic behavior for yourself and others are always valuable.

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Answer: Blaze - 13/06/2012 00:31:24
 T minus 1.5 hours to the experiment

Blaze


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Answer: Luis Gonzalez - 13/06/2012 02:38:20
 Hi Blaze,
Looks like Harry has finally figured it out too.
The drop tells you how much energy is stored (to give back on impact), though some will dissipate.
Looking forward to your results.
I hope you are able to repeat it, as experiments can be affected by unforseen variables.

Regards,
Luis G


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Answer: Blaze - 13/06/2012 04:21:56
 The experiment is complete. It took longer than anticipated due to some unforeseen difficulties as well as intermittent rain. We got a few good videos both from above and from the side. I will see if I can make some of these available to review for those of you who are interested.

Unfortunately I was unable to get the original test apparatus to precess smoothly enough for the experiment (the unforeseen difficulties) so I almost cancelled the test. After some discussion we decided to use my 27 inch bicycle wheel (outside diameter) on a 24 inch arm. The drop was 6 feet. The precession (orbital) velocity was only about 2.4 feet per second which was about half of what I wanted. In all tests we let it precess 720 degrees before dropping it. We had a piece of white panel board on the ground marked every 6 inches so that we would have something to “scale” against from the top view video.

We did a total of 5 or 6 tests since the thing didn’t self destruct on impact. The landing wasn't nearly as violent as I was worried it would be with the wheel basically just bouncing a bit and then rolling away. It did however bend the axle a bit by time we were done testing. I don’t actually remember how many test we did but we have them all on video so I can count them up. The gentlemen I had helping me have the videos and will be emailing them to me tomorrow but we did look at them all tonight. I will study them further over the next few days but the results, although they varied from test to test, are clear.

In every test the wheel stops precessing the instant it is dropped (as expected) and it also continues to move in a linear fashion as it is falling. In the best test the wheel moved linearly about 18 inches from where it was dropped before it hit the ground.

I am happy with the tests as I believe it proves what I have theorized and explained previously in this thread.

Best to all,
Blaze


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Answer: Glenn Hawkins - 13/06/2012 13:05:32
 Good for you Blaze.
So far as I know there were only two, or three people in the world that believed it would stop and drop. Now it will be inferred that only the dead weight in the wheel slung itself away. The experiment needed to be done and it was fun and interesting for everyone. Still it is less equivalent in importance to the collisions. I speak of the one you did with so much power and visual and of my two weaker, less controlled ones. Now that the all these are finished and over, I don't know why we ever though they were necessary. But they were. So much debate and intelligent argument sometimes challenged our thoughts and iron clad certainties. The mind is a strange place. We who knew better still needed the tests.

It is time to look at the pivot.
The action and force of the pivot during acceleration may not actually become a detriment to creating one directional thrusts----- no mater how it behaves. The question may not make a difference in performance. Yet that has always been the prime question and you brought it up. Sometimes the condition is hard to be sure about for some of us. Others among us feel sure, but the have no proof either, except that they believe that the laws of physics cannot be manipulated in a multiple series of ways to create a condition that acts contrary to them. The study of the pivot is the study of the center of gravity.

In all this time the prime question remains disputed. Dose or dose not precession in all set-ups attempt to balance in the center of gravity?

Glenn,

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Answer: Blaze - 13/06/2012 19:57:15
 I have been reviewing the overhead videos over lunch hour and they reveal better distance than I initially thought. There were a total of 6 tests with distances varying between approximately 21 and 26 inches in linear travel which to my mind is fairly consistent considering we were just spinning up the wheel with another 10 inch wheel on a drill.

The reason for the 720 degrees of precession before the drop was to allow the pivot to stabilize before the drop and that did help quite a bit for most of the tests. In all cases, there is no "tumbling" of the wheel. The arm drops with the wheel and it takes very little if any noticeable time to accelerate up the the wheels linear speed. The arm also stays horizontal during the drop until the wheel hits the ground.

I haven't had a chance to see the videos from the side yet. I believe that we only have three from the side.

cheers,
Blaze

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Answer: Luis Gonzalez - 13/06/2012 20:12:02
 Great Experiment Blaze,
The Much Opposed theoretical precept is now backed by basic experimental proof, that anyone can repeat to fortify the theory or to dispel their doubts.

The measurements gathered from your experiments also provide quantification for the energy stored in the gyro’s ORBITING motion during what we have come to call “The Drop” (not to be confused with dropping the gyro from your tree house).

The axel/arm’s DEADWEIGHT Contributions or Detractions to the NET LINEAR motion results should in-fact CANCEL OUT.
Why?... Because one end of the deadweight arm is traveling at the velocity of precession (which may be said to PUSH the resulting linear distance) and the other end of the arm is at virtual stand-still (which may be said to DRAG on the resulting linear distance). Thus averaging the net deadweight effect to zero Linear Distance.

One thing I am interested to know is the nature of angular skewing caused by the dual influences (Push & Drag) from the Arm’s deadweight. I am looking forward to your valuable videos.

Best Regards,
Luis G

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Answer: Glenn Hawkins - 13/06/2012 21:14:13
 wonderful Blaze. Excellent Luis.
Glenn

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Answer: Momentus - 14/06/2012 10:44:32
 Hi Blaze,

….. With a gyroscope the damn thing never does what you expected. Momentus…..

Yet as always it is the result of the experiment which is the sole arbiter of truth.
So yes I got it very wrong in saying it would stop and drop.

Starting with a 50 - 50 compromise, a mass that is half normal matter and half Pixie dust, travelling at say 1 m/sec. When released the normal mass will continue forward at the same 1m/sec whilst falling with an acceleration of 1g. After one second it hits the ground, having fallen 4.9m and travelled forward 1m.

The Pixie dust has no inertial properties, it does nothing, just tags along with the normal mass. Change the proportions to 90% normal matter or 90% Pixie dust and you get the same forward motion over the same time.

I just wish that I had thought this through before the experiment, then I would have appeared wise before the event.

A final thought. No momentum is shorthand for “as us pedants like to say, the dynamic component of a spinning disk does not exhibit linear momentum.”

The design of this experiment does not give any insight as to why this is so.

Momentus


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Answer: Blaze - 14/06/2012 19:24:24
 Hi Momentus. If I correctly understand you, what you are saying is that only the non spinning mass caused the linear motion in the experiment, the spinning mass had no linear momentum and was only pulled along by the non spinning mass.

That would mean that if I redid the experiment (and let's go really big this time) with a 1 ton spinning mass moving at 10 feet per second where ONLY the spinning mass was disconnected from the arm (everything that was dissconnected was spinning, there isn't one bit of mass that is not spinning), the 1 ton spinning mass would just stop and drop? Keeping in mind that there would be no forces acting on the 1 ton spinning mass in a direction opposite to its direction of travel and the 1 ton spinning mass would not be able to act against anything because it is disconnected from everything, it would still just stop and drop?

If I have understood you correctly and that is what you think, then I respectfully disagree.

Blaze

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Answer: Blaze - 15/06/2012 02:26:33
 You can download one of the raw, unedited videos from the experiment from this link
http://min.us/mbcf5A7xWb

The video is 26 seconds long and 34 MB in size and it takes a while to download. The website is free use so I believe they throttle the download to try to entice you to sign up. All the top view videos were good. The videos from the side were almost useless. If the arm wasn’t pointed directly at the side camera when released, you couldn’t see much movement but from the top you can see everything no matter when the gyro was released.

If you download the video you will see a top view video from the edge of my tree house. You will see us spinning up the bicycle wheel, then the wheel precessing a little over 720 degrees, then the string holding the gyro being dropped and the wheel rolling away on the ground. When doing this experiment we fed the string through a piece of aluminum tubing and dropped the tubing and the string at the same time. The idea of the tubing was to help quickly stabilize the pivot string and minimize any coning of the string while the gyro is precessing and it did help. The string holding the pivot settled down quite a bit by time it was let go.

There is a white board on the grass below the gyro that has markings every 6 inches. When the gyro is dropped, it starts moving linearly before it is directly over the board and goes a little more than 3 markings on the board for a total of about 24 or 25 inches of travel even though the gentlemen in the video clearly say “that was about 10 inches”. The total free fall time is only 0.63 seconds so it was very hard to tell exactly how far the gyro traveled when the actual experiment took place. When reviewing the video frame by frame you can see and measure everything.

The camera used runs at 30 frames per second. Hopefully you can look at the video frame by frame from this download. When I look at it frame by frame I can actually see 3 frames right at the beginning of the free fall where the gyro precession is coming to a stop. At 30 frames per second this is actually very close to the calculated amount of time it would take for the precession to stop from this precession speed. So this also proves that it takes some time for precession to stop, it is not instantaneous.

You can also see the pivot end of the arm accelerating very quickly up to the same speed as the wheel is moving. The instant the gentleman starts leaving go of the string and tubing, they are being pulled sideways by the arm as the gyro is falling. There is no reaction at the pivot (ie: pivot does not move at all in the opposite direction that the wheel travels). So this also proves that releasing a gyro from the pivot does not “kick” the pivot in the opposite direction, there is no reaction. This makes sense since the wheel is coasting when in steady state precession. There is nothing driving it so there shouldn’t be any reaction when releasing it.

So that is it for now. Enjoy.

Best to all,
Blaze


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Answer: Glenn Hawkins - 15/06/2012 03:29:17
 It is difficult to view, until you use the stop frame button on the lower left. Then the action and becomes unmistakable proof.

I am very pleased for you, Blaze.


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Answer: Blaze - 15/06/2012 05:06:24
 Thank you Glenn. It came out better than expected. I only expected to prove one thing and ended up proving three.

cheers,
Blaze

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Answer: Momentus - 15/06/2012 12:00:13
 Hi Blaze,

I can only repeat what I said before.
So yes I got it very wrong in saying it would stop and drop.
You are correct to say that as there are no forces trying to stop it, then it will just carry on doing what it was doing. No stop, just drop.

I thought I made it clear that I had completely accepted the evidence of your experiment.

I agree with your conclusion that a ton of pure spinning pixie dust would continue to move at 10ft/sec when released. Size would make no difference.

The point I am trying to convey to you is that the “10ft/sec” motion is not due to inertia, or momentum, or any physical attribute currently known to science.

The freefall drop does not give you any information as to the nature of this mechanism. You see the same thing regardless of what is happening.

The angular motion of a precessing gyroscope starts and stops instantly.
Classical science has not got the known angular motions of gyroscopes wrong. The text book formulae are accurate, the logic and math impeccable. It is only the offset motion of the gyroscope that is not examined and cannot be explained.

If you saw the angular movement of the gyroscope slow down over 3 frames, look for a reason which acknowledges the instantaneous nature of precession/couple.

“Kick at the pivot” Not sure that I have grasped that one. The shaft has angular momentum as it is turning until released. To stop that the gyroscope will apply a reactive torque which will have a precession about the horizontal axis which you will only see in the side view. Any displacement in the plan view will be due to force in the side view. Since the shaft is lightweight there will not be much to see.


Keep up the good work.

Momentus


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Answer: Luis Gonzalez - 15/06/2012 16:58:13
 Blaze,
Your video of the “June 13 Experiment” is unmistakable unambiguous proof.
The “June 13 Experiment” is a noteworthy milestone that adds intellectual capital to the subject matter published in this forum.

We want more, we want more, we want more …we… more…more ….

Congratulations on rare excellence.

My Best Regards,
Luis G

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Answer: Harry K. - 15/06/2012 18:14:23
 Well done, Blaze!

Simple experiments telling us more than 1000 discussions.... ;-)
Luis is right, we want more!

Best regards,
Harry

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Answer: Nitro - 15/06/2012 20:24:39
 Dear Blaze,

With going on sixty replies to this post you have certainly got the wires humming. However I must have missed something amongst it all as I cannot fathom what all of the three things were that you set out to prove. Maybe I’m not just dyslexic, but thick. I guess I’ll have to wait for a sight of the side view videos to help – please hurry and let us know when they can be viewed – to find what I’ve missed and (if I must) I’ll happily eat humble pie while rearranging my understanding – who says men can’t multitask? Your empirical, practical testing of theory is the only way to good progress science and understanding. Lovely to see your simple test show precession end instantly with no further gyro rotation when the axial angle changing force (gravity) is removed on release. If a picture says a thousand words a video says ten thousand – keep up the good work, gabby!? That is not really to call you gabby, more like a variation on the Shakespearian theme of ;- "if music be the food of love; get me a piano sandwich”!

Kind regards
NM



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Answer: Blaze - 16/06/2012 02:17:10
 Gentlemen, thank you all for your kind comments. The videos certainly were important for reviewing and seeing what was really happening. I shall try to get videos for any experiments that I do from now on.

Momentus, I misunderstood what you said previously. I think I understand what you mean now.

Harry, I only set out to prove one thing with the experiment but upon review of the videos I discovered that I proved three things (at least I believe I did, not everyone may agree with me).

FIRST is that when a precessing gyro is released (allowed to free fall in the experiment) the gyro will continue in a linearly while falling. The video clearly shows this.

SECOND, it takes time for precession to start or come to a stop (this is the part that some people don’t agree with). I am talking about “precession motion” which I define as the rotation of the wheel about a vertical axis through its center, NOT the orbital motion of the wheel going around the pivot. A frame by frame viewing of the video shows that the precession motion takes about 3 frames to stop which is very close to the calculated amount of time it would take to stop for the speed that the system is traveling in the video. In a different thread I elaborated on how to calculate the Drop time of an overhung gyro when starting it and the Stop time will be the same as the Drop time. Seeing the precession coming to a stop eluded me the first few times I looked at the video because the pivot is being accelerated to the left at the same time as the precession motion is coming to a stop. Of course, the reason it takes some short period of time for precession to start or stop is because even though the forces act instantly, it still takes time for that given force to change a given speed or rotation of the system. Again, not everyone agrees with me on this one.

THIRD, there is no reaction at the pivot in the opposite direction when the gyro is released. I don’t know if this idea was ever discussed on the forum but I have argued about the idea with several people face to face at work. I said there would be no opposite reaction at the pivot when the gyro is released and they said there would be. Their idea is that if you in space and were swinging a rock around on a string and let the string go, the rock would go in one direction and you would go in the other direction. In this case your hand is the pivot. You can see in the video that there is not even a hint of opposite reaction at the pivot when the gyro is released. I believe this is because the gyro is just coasting when in steady state precession. There is no force causing it to move so there should be no force when it is released. It is kind of like a block of ice sliding on a smooth sheet of ice. The block of ice is just coasting, nothing is speeding it up or slowing it down. If the block of ice slides off a cliff, the sheet of ice does not experience an opposite reaction.

I hadn’t planned on putting any side view videos up for viewing because they really don’t show much. If the arm wasn’t pointed directly at the camera when the gyro was released you really can’t see much movement because the gyro is mostly moving towards or away from the camera.

By the way, can anyone guess which person is me?

Cheers,
Blaze


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Answer: Blaze - 16/06/2012 03:16:49
 I guess I can't read. Nitro was asking what were the three things, not Harry. Sorry about that gents.

Blaze

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Answer: Nitro - 16/06/2012 05:32:14
 Dear friends across the pond and sleeping Britains,

I don't mind being called Harry, Blaze, I've been called worse! Harry may, however, take offence.

It just goes to show that the more we know the more we need to know. While Blaze's video clearly shows that rotation (let’s for clarity call the spin of the gyro “spin” and the rotation caused by precession “rotation”), as expected, ceases the moment the torque caused by gravity is removed (apart, for you pedants, from a tiny amount of movement caused, I say, by insufficient spin in relation to the axial angle changing force – see earlier posts ad inf.) by the string’s release (by the short sighted golden haired Blaze, I would guess) it also shows another monumentally important fact that I and other old timers have been banging on about ( also ad inf.), namely:- that rotary displacement can be converted to linear displacement by the application of gyro dynamics.

That is show in Blaze's video:- without any opposite rotational reaction on the rotational axis (the mass is after all released and therefore disconnected and no longer able to create opposite rotational reaction on the axis) the assembly (gyro, arm etc.) moves linearly –not rotationally – upon its release and its impact (and reaction/s) is/are therefore linear not torsional. This is important so please try to get a grip on it.

It’s now quarter past five in the morning and I think my tiredness may now drown out my tinnitus and allow me to sleep at last, so goodnight and God (or Newton, or whoever is in charge these days) bless.

Kind regards
NM

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Answer: Blaze - 16/06/2012 06:20:39
 "that rotary displacement can be converted to linear displacement by the application of gyro dynamics"

You got it Nitro. There is no other known way of doing that without an opposite reaction and actually very, very few know about this way of doing it. This is ONE of the secrets and a big part of what is required.

Good guess on which one is me, but wrong. The name of the young gentleman releasing the gyro is Trent. He is a technologist that I work with and the person that I have had the most discussions with about what really happens with overhung gyros compared to what classical physics says happens with gyros. He started as everyone starts, with the classical physics view, but he is learning, slowly.

Best to all,
Blaze

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Answer: Ram Firestone - 16/06/2012 17:07:13
 "THIRD, there is no reaction at the pivot in the opposite direction when the gyro is released."

Your gyroscope is already precessing around a fixed point before you drop it. If your tree had been free to slide about, it would have been circling opposite of your gyroscope albeit very slightly since it’s so massive by comparison. After the release it would have been sliding back ever so slightly in the opposite direction.

I can make a similar claim as the one you made above just throwing a baseball. Where is the opposite reaction when I release it? The reaction is during the wind up before I release the ball. It is resisted by the friction between my feet and the ground. If I threw that baseball while floating in space the situation would be entirely different. I would float slowly backwards. In fact I would start to float backwards as my arm went forward. The reaction of your precessing gyroscope is no different. Your pivot is fixed to a tree and by extension fixed to the earth so it does not move.

Your experiment is good for learning but you haven’t proven anything that defies Newton’s third law yet. I said the experiment was hard to do properly. This is why. You are making the same mistake as is made with most gyroscopic propulsion experiments. You are ignoring the fact that your pivot is fixed to the earth in some way or there is significant friction between your pivot and earth. Whenever I point out that the pivot does indeed move when the same thing on a dry ice puck or air table, I get beaten down with nonsense arguments. Unfortunately that doesn’t change the reality of the situation.


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Answer: Glenn Hawkins - 16/06/2012 19:34:38
 Dear Ram,
If you will just hang a string from a table attached to a gyroscope, lie down on the table for a bird's eye view of the hanging string from above, You will see the precession increasing it orbital circumference far out from the table anchor. You will not see a misalignment of the string, as you would from your bird's eye view, if there were one.

There are probably better ways to show the center of gravity is not under the gyroscope, but how can you not understand this as it is?

Best Regards Glenn,

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Answer: Glenn Hawkins - 16/06/2012 19:39:38
 Excuse my misstatement, Torque displaces the weight of the gyro to the pivot as so many know. Why is it you cannot realize that?

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Answer: Blaze - 17/06/2012 00:16:42
 Hi Ram, thanks for your input. You and my buddy Trent should really have a talk (the young gentleman in the video releasing the gyro). You would get along famously because you both say the same things. You may both be correct too in some cases. What you are talking about is a barycenter, “the weighted average location of all the mass in a body or group of bodies”.

So, let us first talk about the perfect gyro.

When a perfect gyro (no friction of any kind and no non-spinning mass) is in steady state precession there is no sideways (horizontal) force on the pivot. The precession speed is not increasing or decreasing therefore there is no force (or force couple) involved in moving the gyro around the pivot because if there were, then the gyro precession speed would have to be increasing or decreasing (there is of course still the force of gravity pulling down but that is not in the precession plain). The gyro is just “coasting” like much like the block of ice sliding on a smooth sheet of ice except that the gyro is going in a circle rather than in a straight line. I believe it is this “coasting” concept you are having difficulty with, the idea that there is no horizontal force on the pivot during steady state precession. Again, if there were a force couple during steady state precession, then precession would not be steady state, the gyro precession would be either speeding up or slowing down.

Since there are no forces in the horizontal direction acting on the pivot during steady state precession, there can be no opposite reaction when the gyro is released because there is no force to react against. In this scenario if there was a reaction at the pivot you would now have a unidirectional force (ie: no a force pair).

However, if the gyro precession were speeding up or slowing down, then there are horizontal forces on the pivot and if the pivot were free to move (no friction remember) then you would in fact have a barycenter situation. I have seen this happen briefly many times when starting my Tedco gyros, the pivot moves sideways, therefore there is a horizontal force on the pivot when starting the gyro. I have even seen this when stopping my toy gyros but it is far more difficult to do. If the gyro were released to fly freely through the air while there is a horizontal force on the pivot there would in fact be a reaction at the pivot because of the barycenter situation. However, in steady state precession there is no horizontal force on the pivot so there cannot be any reaction when the gyro is released.

Now let us talk about a real gyro operating in space.

In short you can pretty much forget almost everything I said above. You will probably always have barycenter situation because there is both friction and non spinning mass. Even if you used magnetic bearings in a vacuum I believe you would still have “friction” because of eddy currents, however, that friction would be very small. So while friction can be your friend on Earth, it generally isn’t in space. Anytime you have friction in a rotating system on a spacecraft it has to be equally countered in the opposite direction to eliminate it or the whole spacecraft will begin to rotate, usually in a manner that you don’t want. This means that to make something like a gyro to work in space without unwanted rotations or motions, it will be far more difficult than on Earth, however, I don’t believe it will be impossible. It would take the right combination of some truly excellent engineering and “mirrored” gyro systems to get rid of all those unwanted rotations and motions.

Cheers,
Blaze


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Answer: Glenn Hawkins - 17/06/2012 01:03:40
 Hi Blaze.
The key words are static and dynamic. I don't see a difficulty in engineering for space, but as I insisted the design is necessarily complicated, which I avoid here. Basically the dirt simple is, a minimum of four gyros alined in a square and pushing against each other-- not against the ship. That gives them static stability. The precession of each twin pairs moves opposite to the other pair providing dynamic balance. This may not be explained well, but it is correct and I am sort of in a hurry.
Cheers, Have a good Evening,
Glenn

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Answer: Blaze - 17/06/2012 01:17:47
 Hi Glenn. I agree, that is exactly what I thought too, 4 gyro systems to eliminate all the unwanted rotations and motions. These 4 systems would have to be perfectly synchronized though.

Cheers,
Blaze

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Answer: Ram Firestone - 17/06/2012 06:22:18
 “When a perfect gyro (no friction of any kind and no non-spinning mass) is in steady state precession there is no sideways (horizontal) force on the pivot.”

This is the first point I disagree with. As I said before I have spun custom precessing tops on air table pucks. The puck (i.e the pivot) still moves and moves A LOT since the top is far more massive than the puck. I see no evidence that points to a theory that if you were to remove every last bit of friction from the system the pivot would suddenly not move. In fact what is observable is the exact opposite. The more frication that is removed the more Newtonian the behavior is. However in your case you fixed the pivot and as far as I can tell your experiment did exactly what might be expected under those circumstances.

“I believe it is this “coasting” concept you are having difficulty with, the idea that there is no horizontal force on the pivot during steady state precession.”

Well, I am having difficulty with the idea there is no horizontal force on the pivot or rather I simply don’t believe it is in fact true. As for coasting, yes I think that’s a good analogy since no energy is being used in the theoretical perfect experiment. It’s just like spinning a ball in space. Note however the spinning ball spins around its center of mass. Your coasting analogy actually works against you. If no energy is being used no mass can be accelerated.

“However, in steady state precession there is no horizontal force on the pivot so there cannot be any reaction when the gyro is released.”

Again we have this repeated claim that has never been proven and directly contradicts Newton. First off I don’t see why a change in precession speed is necessary for a horizontal force on the pivot. I’m just going to ignore that for now since I don’t understand it and just say that in every experiment I’ve ever seen where friction is minimized the pivot moves. If you can prove me (and Newton) wrong I’ll be happy to accept it however it DOES need to be proven.


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Answer: Harry K. - 17/06/2012 10:56:29
 Hello Gentlemen,

Blaze´s experiment proves that angular momentum is stored in precession movement of an overhung gyro system, nothing more. It does not prove that precession movement does occur in the centre of spinning mass (with is may conviction). However, if precession occurs in the centre of spinning mass, there must be a reaction force at the pivot, the counter force of the couple caused by precession torque.
Ram is pretty right that the tree and thus the earth provides ths reaction force. Furthermore the precession couple (=torque) disappears fast when the gyro falls down to earth because there is no tilting torque acting. Thus only the stored angular momentum around the centre of spinning mass (not the stored angular momenum in orbital movement!) would cause a reaction force at the pivot, however, this force may be to small to notify in this experimental setup.

I think the main problem for some of us is to consider that any gyro actions as well as any gyro reactions are caused by torques (couples) but not by single forces!
Blaze, you should take Trent more seriously. ;-) Are you the guy how spins up the gyro???

Best regards,
Harry


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Answer: Glenn Hawkins - 17/06/2012 12:39:31
 Dear Ram,
The pivot dose not move of it's on accord. The flywheel pulls the pivot outward toward it. The string does not show a pendulum like angle forward, or rearward to the direction of precession. It shows only the outward pendulum like angle, which is a position quite removed from the center of gravity.

Respectfully, I have never accepted your top/puck experiment. The reasons have been given to no avail. I am sorry.

Yours Truly,
Glenn

////////////////////////
Dear Harry,
I actually know how to provide inarguable proof, normally, but I don't know how to quench this argument as I predict inarguable proof is useless and would not make a difference here among us. ; -)

Respectfully Yours,
Glenn

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Answer: Blaze - 17/06/2012 15:09:20
 Hi Ram.
“Your coasting analogy actually works against you. If no energy is being used no mass can be accelerated.”

Ah yes, but that is exactly the point. The mass is NOT being accelerated (or decelerated) when in the coasting mode (steady state precession).

“why a change in precession speed is necessary for a horizontal force on the pivot” is because then mass is being accelerated or decelerated and a torque couple arises when this happens, therefore a horizontal force on the pivot.

Harry, why would you guess I was the guy spinning up the gyro? You are correct.

Happy Fathers Day to all you fathers out there,
Blaze



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Answer: Ram Firestone - 17/06/2012 16:08:54
 "Ah yes, but that is exactly the point. The mass is NOT being accelerated (or decelerated) when in the coasting mode (steady state precession). "

But it is, if it is not precessing around its center of mass. In fact it’s accelerating in all directions within the plane of precession. Let’s however consider this in 2D for a second. Looking at your experiment from the side the center of mass looks like it accelerates from a stop on one side, reaches its maximum velocity in the middle, then decelerates until it stops and reverses on the other side. Rinse and repeat. So it would be under constant acceleration. The reason why this isn’t actually happening is the earth is the other part of the equation. Yes, your experiment can be considered to be moving the earth an infinitesimally small amount. I can do the same thing doing pushups and so can you. Don’t you feel strong now :-)

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Answer: Nitro - 17/06/2012 21:31:48
 Dear Blaze,

It is clear that Ram, like Dr Huge, has set himself up as another high priest of the religion of Newton. No amount of intelligent discussion or presentation of what others consider proof will cause him to waver in, or re-examine, his belief in every word of his “bible”. You can point out that the convocations of bishops at Nicaea rewrote it twice to bring what all the bishops preached into line and that our King James did another clean up job, while translating from the Latin, as well, but every modern word will still, in his mind, be written by his “god” and therefore be immutable (I may have got mixed up with something other than gyros there). I’ve gone back over some of my letters to this forum from 2003 and, way back then, I came to the conclusion that his (Ram’s) seeing the nail holes would not be enough, he would not be happy ‘til he banged the nails in himself. Trouble is he seems prepared to do the talk talk but not the walk walk needed to make the nails himself. Unlike him I, like you Blaze, got my finger out and tested and made and tested and made. Oh! I made many mistakes, and probably still have many misconceptions, but I made the one shot machine to confirm a theory (now in the public domain ) and went on to a make a multishot version. The multishot was overly large in relation to its gyros and its motion was, as has been said, a succession of small impulses (movement and dead stop, movement and dead stop etc.) and therefore did not seem of much use except to prove a point (if it ever got used on a manned space shot it would likely shake all the astronauts’ dental fillings out!). Yes! I know I could have smoothed things out with the gyro equivalent of a V12 but feeding the family became more important so it was abandoned pending more time and money being available. Sadly I can now find no pictures or video of the multishot to put up (I still have bits of its frame and the gyros so when I retire - who knows) but, like the one shot, it pretty comprehensively buggered up the first, second and third laws of the great god Newton – in my ever so humble opinion. I don’t know why I bothered to put IMHO, as, regardless of such riders, all the high priests will already be running around screaming heresy and demanding that the existence of a different god be proved by means only acceptable to them for fear of loosing their power of didactic argument.

Blaze, I am not particularly interested in trying to satisfy Ram’s whims (indeed I don’t any longer think that Ram is interested in satisfying his whims – or he would surely have devised his own tests and put them up for examination by now) but you may, for your own interest, like to carry out a further simple test with the set up you have shown on your video:-

With the end of the arm suspended by the string (preferably made longer and clear of obstruction) from its upper fixed point to clearly show any reaction at its attachment point to the arm that acts as the pivot point) spin up and release the gyro (but not the string) and video from above as before. If Ram and others are right there should be displacement of the string in the opposite direction to the gyro’s rotational displacement. You might also like to accurately ascertain with a plumb bob, the point where the dead vertical string release point should be and mark it with a rod sticking up vertically from the ground. The fixed rod would give a good visual indication of any changes in the pivot point – and there will be changes, but not in the opposite orbital direction to the gyro orbit that Ram thinks. You could also use the same starting set up but with the gyro end of the arm suspended by a thread to be burnt through with a small blow lamp (pencil flame) to prevent human input to the motions. This second test could be used, with a horizontal camera position, aligned with the gyro’s axis, to ascertain and quantify any gyro drop at release, accurately.

A video of a gyro suspended by a string has been put up by me, in the past, that clearly shows that Ram’s opposite orbital reaction is not only missing but that the axis’ movement is in the wrong, outward, direction following the gyro’s orbit, likely caused by the centrifugal effect of the rotating non gyrodynamic mass. Needless to say Ram found a way to convince himself that his opposite reaction was there but that it was masked by the colour of the string/Magnus effect/Mesons/the phase of the moon/lack of bichir fish etc. etc.

Kind regards
NM


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Answer: Luis Gonzalez - 18/06/2012 00:45:03
 Hi All,
It’s true that the old timers have, for close to 40 years, discussed a central point that Blaze has demonstrated with a simple and elegant experiment (whether precession’s motion has equal-and-opposite reaction, or not).

The old timers never sufficiently parsed the problem-space to determine which “states” of precession have (and which do not have) equal and opposite reaction, and unfortunately most do not see the difference nor spend sufficient effort to understanding it – Hi Ram.
(Notwithstanding, the explanation has been posted in this forum for years (give or take) but appears to have gone completely ignored.)

It is indeed Noteworthy that after nearly 40 years of failure to prove a point, which is central to this forum’s quest; here we have an experimental “design” capable of proving this very point for the first time (and it only took 9 days from inception to execution of the tests).
With small refinements this experiment should make all its clams irrefutable.

Question:
Will someone explain how Blaze’s gyro pushes against the tree when it is released to starts its linear journey??
Frankly, I am perplexed…
(1) Upon release the gyro moves LEFT with a momentum of 7.2+ foot-pounds per second, and (2) somehow pushes on the cord and/or the aluminum tubing (how?).
(3) The cord and tube do NOT move to the RIGHT (opposite of gyro motion) but somehow manage to push against the tree… with a force sufficient to generate an opposite momentum of 7.2+ foot-pounds per second.
How does the cord and aluminum tube apply this force against the tree (Considering that both are dragged along by the momentum of the gyro)?

I look forward to a sensible clear explanation; otherwise I think this is the best experiment in support of the old timers’ 40 year claim.
I can’t wait to see a good response.

Regards,
Luis G

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Answer: Glenn Hawkins - 18/06/2012 04:28:41
 Hi Luis,
I am with you. Blaze deserves praise and acknowledgements. I am proud of him. Though he is young he is one of the sharpest people on here, not because of the test, but because of what he understood and related before the test. The test confirms what he knew perfectly well, but it benefited others and proved to them he was correct. I think that is right. Correct me If I am wrong.

I knew too, I think you knew as well mostly what the outcomes would be. These old timer discussions? It was a case only of some of us not knowing, while others knew. The point did not need to be proven to the ones who knew. There was no question. It was settled for them.

You ask, “Will someone explain how Blaze’s gyro pushes against the tree when it is released to starts its linear journey??
Frankly, I am perplexed…”

Yes. I will. It has been discussed. Blaze knows. Look at the video with stop frame when the gyro is released. You will see the cord jerk out of sight under the tree in the opposite direction to precession. The jerk and cord displacement last only a fraction of a second.

The reason: The drop, the initial acceleration downward is over and done with so quickly, but during that time when torque is being twisted to carry on in the horizontal plain and there is equal and opposite reaction. Once this supper quick torsion is over and done with, the gyro coast. It is in a steady state mode and in this mode there is no equal and opposite reaction. For just a fleeting instant precession pushed against the tree, but not afterwards.

Luis, I think Blaze may tell you the same thing in different words. Good question.

Best Regards,
Glenn

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Answer: Ram Firestone - 18/06/2012 06:37:00
 I would suggest blaze, you keep doing your experiments and find your own answers. But please keep an open mind and try to see what’s actually there and not what you think you want to see. Otherwise you will end up 40 years from now as an angry “true believer” with nothing to actually show for it. I see no reason why my disagreement with the interpretation of any of these results should insight such a strong reaction, but there you have it. Consider what your buddy Trent is telling you. He sounds like a sharp guy. If you really want to pursue this I would concentrate on the drop phase because there is no energy to be had during steady state precession.

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Answer: Harry K. - 18/06/2012 11:54:18
 Hello Luis,

"Will someone explain how Blaze’s gyro pushes against the tree when it is released to starts its linear journey??
Frankly, I am perplexed… "

Ask the young guy Trent how much force he needed to hold the overhung gyro in position during precession. The force to hold the gyro in position is provided at least by the earth. As soon as he released the alu tube and string, the by earth provided counter force acts in counter direction. Only because this fact is not noticable in the video, it does not meaning it is not present.

So far I know is Ram pretty right in anything what he stated in his last postings. It`s your problem if you feel not happy with it.
I`m perplexed too, that you didn't get it yet.

Regards,
Harry

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Answer: Glenn Hawkins - 18/06/2012 12:58:45
 Dear Harry K.
The test was conducted during freefall. Before and after has no bearing. The backdrop of the earth as a resistance was removed. Gravity was not in play. It is what it is.

I will go on to other things. Have a great day and I am looking forward to seeing more of your very often advanced knowledge in future post.
Glenn,

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Answer: Harry K. - 18/06/2012 13:27:14
 Dear Glenn,

The point is the exactly moment when the pivot is released. In this moment the backdrop of the earth is still acting.

Harry

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Answer: Luis Gonzalez - 18/06/2012 16:36:42
 Hmmm…
Looks like someone saw 7.2+ foot pounds per second of “EQUAL” and “OPPOSITE” momentum against the string.
Someone else saw nothing but told me that I have a problem and am not happy about something… (I will seek therapy immediately on this advice).

I suppose we are all trying our best…

…We are talking about 7.2+ foot pounds per second of “EQUAL”….. and “OPPOSITE”…. (horizontal and to the right…) momentum applied against the tree, not against a rope as it falls through the air.
It need to be equal and opposite… got it?

My explanation is that the opposite momentum was NOT “EQUAL” i.e., not 7.2+ foot pounds per second.

Regards,
Luis G

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Answer: Harry K. - 18/06/2012 20:02:09
 Dear angry Luis,

First, I hate it to do not use the standard SI-system.
You state "7.2+ foot pounds per second of “EQUAL” and “OPPOSITE” momentum against the string". From where did you get this data?

If got the following information about technical data of this experiment from this thread:
- diameter of the wheel = 12 inch = 0,3048 meter
- weight of the wheel = 3 pounds = 1,360 kg
- pivot arm length = 24 inch = 0,61 meter
- weight of pivot arm = 0,5 pounds = 0,227 kg
I have measured the precession speed which is about 4,5 seconds for one revolution.

Now let's do math:
Number of revolutions - precession:
4,5 seconds per revolution = 0,23 revolutions per seconf
Angular velocity:
W = 2 x Pi x n -> 2 x 3,14 x 0,23 = 1,45 1/s
Moment of inertia - gyro
J = m x r x r -> 1,36kg x 0,61m x 0,61m = 0,5 kg x m x m
Angular momentum of pivoting mass - gyro:
L = J X W -> 0.5 kg x m x m x 0,23 = 0,725 Nm
Centrifugal force - gyro:
Fz = m x r x W x W -> 1,360 kg x 0,61 x 1,45 x 1,45 = 1,74 N

Moment of inertia - pivot arm
J = m x r x r -> 0,227kg x 0,305m x 0,305m = 0,021 kg x m x m
Angular momentum of pivoting mass - pivot arm:
L = J X W -> 0.021 kg x m x m x 0,23 = 0,00483 Nm
Centrifugal force - pivot arm:
Fz = m x r x W x W -> 0,227 kg x 0,305 x 1,45 x 1,45 = 0,145 N

Complete acting angular momentum:
Lc = 0,725 Nm + 0,00483 Nm = 0,72983 Nm

Complete acting centrifugal force:
Fzc = 1,74 N + 0,145 N = 1,885 N

Tilting torque, which causes the precession movement:
Mt = (13,3 N x 061 m) + (2,2 N x 0,305 m) = 8,78 Nm

Your stated 7,2 foot pounds per second is equivalent to 9,76 Nm which is more than 13
times more than calculated! - Or is my math wrong?

Also the angular momentum does not act "per second" but is stored in precession movement in form of energy. How long it takes to convert this energy can be calculated as well.

Please let me know if you find any mistakes in my calculations.

Kind regards,
Harry




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Answer: Luis Gonzalez - 18/06/2012 21:20:15
 Great work Harry,
Thank you.
As usual I will take time to verify your premises and calculations.
You know this will be a bit slower than just writing English for me, as I am permanently involved in multiple projects.
Best Regards,
Luis G

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Answer: Luis Gonzalez - 19/06/2012 17:45:42
 Harry,
The 3 pound flywheel moved 18 inches (1.5 feet) horizontally to the left in the video, during the same time that it fell 6 feet to the ground.

Therefore the horizontal motion to the left (in the video) occurred at a velocity of 2.45 feet per second, which yields a “Linear Momentum” of 7.35 Foot-Pounds per Second.
Simple!

The purpose of this experiment does not require angular considerations to accurately calculate NET RESULTS (except for the sole enjoyment of math etc).

I am through with this thread for now, and moving on to other things.

Best Regards,
Luis G

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Answer: Harry K. - 19/06/2012 19:27:32
 Hello Luis,

Based on your stated data (not prooved by myself) I have calculated this:

Flywheel, 3 pounds = 1,36 kg moved 18 inches = 0,4572 m to the left side with a velocity of 2,45 feet per second = 0,747 m/s

Kinetic energy:
Ekin = 1/2 * m * v * v = 1/2 * 1,36 * 0,747 * 0,747 = 0,379 Nm = 0,279 Foot pound
Also simple !

Either my math or your math is wrong...

Best regards,
Harry

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Answer: Ram Firestone - 19/06/2012 19:30:30
 OK well just using your numbers:

K = 1/2 * m * v^2

So:

K = 1/2 * 1.36077711kg * (0.74676m/s ^ 2)

So:

K = 0.37941901 j = 0.279845101 ft-lbs

There is no "per second". It’s a measure of energy not power.


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Answer: Ram Firestone - 19/06/2012 19:32:09
 Bahhh! You beat me to it :-) That's what I get for making coffee in the middle of posting.

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Answer: Harry K. - 19/06/2012 21:03:10
 Hi Ram,

Sorry... :-)
I'm glad to see that you use the same math.

Have a nice evening,
Harry

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Answer: Luis Gonzalez - 20/06/2012 01:11:08
 You guys are geniuses but we are talking apples versus oranges.
I chose to calculate “LINEAR MOMENTUM” and you guys are calculating KINETIC ENERGY.
Do you geniuses think they are the same thing?
I am Open for advice…

Regards,
Luis G

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Answer: Blaze - 20/06/2012 01:41:59
 So if it is linear momentum we are talking about here then
momentum = mv

The actual gyro used was not the one I expected to use. This one was bigger. The wheel diameter was 27 inches and the whole thing weighed 6.5 pounds. I checked the frame numbers for the last 360 degrees and got a velocity of 3.02 feet per second.

So converting to metric, this would be
momentum = 2.95 x 0.92 = 2.71 kg-m/s

Best to all,
Blaze

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Answer: Ram Firestone - 20/06/2012 04:50:52
 Louis, I guess you did say “linear momentum” but you started giving things in foot pounds or rather foot ponds per second which would be a unit of power so I’m not exactly sure what you were trying for. Exactly what are you trying to prove here anyway? The whole idea behind momentum is that is that it’s always conserved. Now you are claiming it’s not always conserved, so why even talk about momentum? Since its conveniently impossible the measure the movement of the earth caused by this experiment I guess you can claim anything you want until the same experiment is done on something that has minimal friction to the earth. But then you guys tend to argue that such experiments are no good (for what reason I still don’t know) so again you never have to prove anything. No wonder this has gone on for “40 years” with nothing to show for it. Any experiment that would disprove your theory is conveniently avoided or discounted.

Let’s look at this your way:

p = m * v

so:

p = 3 lbs * 2.45 ft/s
or
p = 1.36077711kg * 0.74676m/s

so:

p = 7.35 lbs * ft/s
or:
p = 1.016173 Kg*m/s

So far so good, but now by conservation of momentum the earth should be moving

1.70093569E-25 m/s

in the opposite direction, which is impossible to measure. If you are so sure momentum is not conserved in this case why not do the experiment on something you can measure rather than the earth?


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Answer: Nitro - 20/06/2012 09:14:26
 Dear Blaze,

As usual Ram has ignored the inconvenient truth and resorted to the math created to describe the earlier religion which surprise, surprise, can only serve to confirm the earlier religion. The math to describe the observable anomalies of the new religion shown in your and my videos have yet to be written.

To be able to move the world even the immeasurable amount prescribed by the old math, the article attaching the gyro assembly to the world - namely the string - would have to be straining to be moved in the opposite direction to the movement of the gyro assembly. So if Ram and the mathematical geniuses that are nailed to and by the old religion care to look again at your video, Blaze, they should clearly see at the moment of release, the string straining in the opposite direction to the gyros direction of movement. Oops! Tilt! It doesn’t.

Kind regards
NM
PS Blaze, if you care to use a longer string and release by burning through a support thread (as suggested earlier), to remove the joggling caused by human release, it will make this and other effects clearer to the worshipers in the Pantheon. However, I bet they will still reach for the old equations.


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Answer: Glenn Hawkins - 20/06/2012 11:38:34
 “The definition of insanity is going around in a circle expecting to get somewhere.”
US President Bill Clinton

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Answer: Luis Gonzalez - 20/06/2012 14:27:36
 Hi Nitro,
Your explanation regarding Blaze's experiment assembly and how the cord would have to strain in the opposite direction should drive the point home even for people who can’t understand words written in English by someone who thinks in Spanish.

Regarding the math, I don’t think we will need to throw away the existing paradigms but rather finesse and/or expand some of the parameters (similar to the way relativity tweaked the rate at which time and motion flow).

Some genius is going to come up with some sort of spin-constant or spin-variable that reconciles the seemingly disparate equations.

Best Regards,
Luis G

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Answer: Harry K. - 20/06/2012 16:26:59
 Hello Luis,

Ok you want to use momentum (impulse) which does not tell much about stored kinetic energy in precession.
Also you have used a different circumferential speed of 2,45 ft/s = 0,74676 m/s. This means that the precessing gyro needs 5,13 seconds for 1 revolution.
Thus I have to adjust my calculations accordingly:
n = 0,74676 / 2 / Pi / 0,61 = 0,19483695 1/s
W = 2 * Pi * 0,19483695 = 1,22419672 1/s
J = 1,36 * 0,61 * 0,61 = 0,506056 kg*m^2
Angular momentum L = 0,506056 * 1,22419672 = 0,61951209 kg*m^2 /s

Linear momentum: p = m * v = 1,36 * 0,74676 = 1,0155936 kg*m/s

To calculate linear momentum based on angular momentum, angular momentum has to be divided by the radius of the rotating mass:

p = L / r = 0,61951209 / 0,61 = 1,0155987 kg*m^2/s

The very small deviation is caused by rounding errors.

You can see that angular momentum in connection of the radius is equal to linear momentum. However, in a closed system the sum of all momentum is zero. Before the gyro drops to ground, the earth provides counter angular momentum to the angular momentum in precession. And YES, this counter angular momentum is transferred with the string.

In the moment when the string is released, Earth provides linear momentum in counter direction to the linear momentum of the falling gyro mass.
The sum of momentum is in both states zero.

That the string moves with the gyro in the same direction does not wonder, because it is released from earth and thus the earth cannot act at the string anymore. But the earth has still absorbed the counter momentum before the strin was released.

Anyway, some of you may go the easy way and refuse the "old math"... I don´t care. Be happy! ;-)

Harry

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Answer: Nitro - 20/06/2012 16:33:24
 Hi Luis,

You are, of course, right. There is nothing wrong with almost all the uses of the equations that have stood us in good stead for so long. However, as I’ve said before; “equal but not oposite” applies when there are two axes of rotation. All the mass, acceleration, reaction, energy transfer and forces are in there, they are there just not where you would normally expect them.

Kind regards
NM


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Answer: Luis Gonzalez - 20/06/2012 18:02:29
 Hi Nitro,
Equal but not 100% opposite sounds about right.

Best Regards,
Luis G

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Answer: Ram Firestone - 20/06/2012 19:25:12
 "Your explanation regarding Blaze's experiment assembly and how the cord would have to strain in the opposite direction should drive the point home even for people who can’t understand words written in English by someone who thinks in Spanish."

First off the cord is swinging all over the place. Second it has the entire weight of the apparatus trying to strengthen it out. Why you guys think a hanging chord mimics a low friction surface is beyond me. Hang it from an air cart of some sort and it will move. Yes it’s more difficult but if you are so convinced you are right it should be worth it since it will make you rich and famous to disprove the 3rd law of motion.

Just forget the math and do the experiment correctly. Then you will see there is an opposite reaction. You also conveniently ignore my point about conservation of energy. If the center of mass is truly moving in a circle as you claim then it is under constant acceleration which takes energy. Every time I bring this up it somehow gets swept under the rug.


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Answer: Nitro - 21/06/2012 18:19:29
 Dear Blaze,

You will, after reading Ram’s response above, perhaps better understand why I suggested the careful release, by burning through a thread, and using a longer free string. Even though doing this is unnecessary for normal observant people, it will show more clearly the direction of forces acting to change the string's hanging angle (precisely because the weight of the apparatus is trying to straighten (!) it out. An air table doesn’t have that advantage.) this, as I have said before, will not satisfy any fundamentalists, like Ram, but may help others understand what is being shown. I am sure Blaze, like me, you are not ignoring the point made by Ram that the conservation of energy laws would be broken but understand that the conservation laws, while still in there, are different when mass rotation on two axes is involved, so the point is moot (US meaning). The point about the anomalies caused by gyros has been gone over and over many times in different ways over the years in attempts to ease his understanding. After Ram’s complaint that his point was being ignored he surely hasn’t been ignoring these points made before by others? It’s starting to point towards the point that trying to get Ram to understand is pointless!!

Kind regards
NM


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Answer: Ram Firestone - 21/06/2012 19:13:06
 "the conservation laws, while still in there, are different when mass rotation on two axes"

It must be nice to be able to make up rules of physics as you go, without ever having to prove anything.

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Answer: Nitro - 21/06/2012 20:19:35
 Dear blaze,

Don’t bother doing the long string test yourself. I was only enjoying winding up Ram, with suggesting pendulum tests instead of air table tests that he loves, but he seems to have a self winding mechanism. The string demonstration, showing the forces involved, has always been on our own web site right here:-

http://www.gyros.biz/lecture/wmv/9.wmv

As most, except for Ram, of course, will see, there is no sight of opposite rotation of the suspending “string” or sight of the gyro being anywhere near the centre of rotation until the gyro’s spin is slowed to the point (now that is a good point to make) where it is made to show more Newtonian action than gyrodynamic action (see earlier posts of mine and others, ad inf.).

One might want to ask Ram:- Where is the rotation around the centre of mass? Why do you need math before doing an experiment correctly unless it is just to prove the math – surely this is scientifically wrong? Where is the opposite reaction? Conservation of energy is not a problem with orbiting planets so what’s your problem with it here?

Oh! and Ram, I have no need for acknowledgement for disproving the 3rd law, I am already rich, and fame is only caused by the stupid thinking someone else is of more value – who needs adulation from stupid people? And yes, it is nice!

Kind regards
NM


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Answer: Ram Firestone - 21/06/2012 23:55:39
 "One might want to ask Ram:- Where is the rotation around the centre of mass? "

The second test is the only one that matters since it’s not swinging wildly. Notice starting at 4:04 the gyroscope is NOT rotating around the pivot but closer to its center of mass while the string is moving in a circle. There is still a slight pendulum effect in combination with this but you can clearly see this effect despite the centering effect of the string which is much reduced due to its length. Thanks for posting this video and proving my point.


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Answer: Ram Firestone - 22/06/2012 02:06:30
 BTW when a gyroscope is spinning slowly it precesses faster. This gives the string less time to center itself. This is why you tend to see this kind of reaction more with a slower spinning gyroscope. It has nothing to do with dead weight or any other fanciful ideas. Put it on any air table however and you will see it even with a fast spinning gyroscope with slower precession. OH wait! That would prove you wrong. Better not do that test. It’s important to keep the illusion of no opposite reaction so we can go for another 40 years make the same claim.

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Answer: Glenn Hawkins - 23/06/2012 02:17:22
 This discussion has moved almost beyound insanity.

Inertia is a simple idea. Mater whether still or moving resist being moved in any direction. Assume that in free space, the slightest force can move the greatest mass. However, the smaller the force, the greater the mass, the slower the acceleration. In straight line direction a mass that is set to motion drifts forever. In rotation the drift is stopped every time the direction of travel is reversed, which happen every one-half rotation. When a flywheel is being tilted it is constantly being forced out of it comfortable plane of rotation into another plane and yet another and yet another and so on. Rotation can be so fast that force to change the plane of inertial resistance is given almost not time to complete its task. At extreme speed, tilting only a fraction of an inch could take weeks, months or more. More tilting force must be add to overcome the resistance to increase a faster pace of tilting and all that goes with it.

I want you Ram, Harry and (Momentus no dark mater, all is mechanical) to concentrate of the resistance of the fly wheel to tilting. I have just explained why the resistance is there and why it is constantly correcting against a continuous tilting force. We have wrongly termed this resistance as ‘deflections’. Deflections mean to bounce off like a rubber ball bounces. Inertial resistance is not spring loaded like that. Constant resistance is constantly overcome and constantly resets to resist again. Still we can use the term deflecting/resistance if we wish. When the deflection/resistance is overcome by tilting forces, it torques (twist) in opposition to the tilting force. The deflection/resistance torques/twists down on the pivot of an over hung gyroscope. The weight of the gyroscope is transferred to the pivot.

The result of that is that the gyroscope cannot rotate around it’s center of mass, for it’s essences of weight, not real weight, has been moved to the pivot. It must rotate around its essences weight from the after math of torque, and it is imposable for it to rotate around its center of mass in such a manipulated condition as exist.

I can hardly believe you do not see what you see, but if you can’t see and realize, you must go over and over this explanation, until you finally understand it. Don’t argue, think! Study, think, think, study the few paragraphs, think, think and think. It is all hear for you.
you have refuse to read carefully and re-read and think, until you understood. It is like we all do not want to learn, we refuse to learn from another man’s mind. We seem to just want to talk and talk and go on and on and never RESPOND to anything against our preferred ideas.

I’m not cleaning this up. If there is a problem I will come back to fix it.

Glenn of limited sanity himself.


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Answer: Nitro - 23/06/2012 15:27:06
 Dear Blaze,

Its a beautiful sunny day here on the island. The storms of the last few days have abated leaving behind still crashing waves to add the most brilliant white crests highlighting the sea, its blue changed to a translucent turquois by the sun’s light reflecting off the submerged golden sands in the shallows before deepening to the darkest of blues further out in deep water. The bright coloured, clinker built, fishing boats are back out on their moorings in Cobo Bay creating the human interest to this glorious scene and enabling every amateur photographer to appear an expert. My gout has abated with the storms, Newton (or Laithwaite, or your favourite god) is in his heaven and all is right with the world.

To save me having to cover myself in sun cream and a avoid the kind of nose that a Glaswegian alcoholic would be proud of, I have taken a break in the cool of my cathedral ceilinged, granite walled, “old lounge”. I have used this “cooling time” to refresh my memories of a few of the fundamentals of gyrodynamics, so ably shown in the Christmas lectures at the Royal Society (Tossers!) by the late, great, Laithwaite. (yes, he made a few simple errors but let those amongst us who have not, cast the first stone). The videos are worth revisiting from time to time (and thank you “Gyroscope.org Glen” for having created this site and putting up the videos) as discussions on the nature of gyros tend to get bogged down in petty distractions while ignoring the “Elephant in the room” or “the bleeding obvious” as we (adoptive) Cockneys say. I commend to you, Blaze, Video Number 4 from the series on this very website:-

http://www.gyroscopes.org/1974lecture.asp

It shows answers to some of the things you seem to be seeking answers to and:- That there is no observable inertia in a precessing gyro, provided it has little or no non gyroscopic mass, (this is clearly shown here that when a weight causing precession is suddenly added or removed, the gyro’s acceleration or deceleration into or out of precession is, to all intents and purposes, instant). That there is no drop (A gyro does not need to drop or be moved - merely to have a force applied as though to change its axial angle - for an actual axial angle change by precession to be caused). That my description is right (The effect of having too much weight trying to change the axial angle, or the weight having too much leverage or the gyro having too little spin, will progress to nutation). That these three things will also cause an increasing conversion from gyroscopic precession effects to the effects of rotating a non spinning mass (a gyro will therefor increasingly follow Newtonian law, instead of gyrodynamic law, as nutation nears). You should, of course, carry out your own confirming tests.

To carry out successful tests of precessional effects, a gyro must have large mass, and high spin speed in relation to a given force applied to change its axial angle, if Newtonian reactions are not to creep in and sod up any observations.

Kind regards
NM

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Answer: Blaze - 24/06/2012 17:16:10
 Hi Glenn. Good post, however, I don't quite understand what you mean by the last line of the first paragraph.

I agree that it is all mechanical and as such can be explained with simple mechanical motions and movements that combined together add up to the effects we see in gyros. The problem is that there is always more than one of these motions and/or movements happening at the same time so we have to be aware of EVERY motion/movement that is happening to understand the total effect. As I have said before, there is no magic here. There is nothing so mysterious that can't be explained with the proper combination motions, however, the total result can look quite mysterious at times.

To all reading this:
What was I trying to prove here? Simply that the gyro will continue in a straight path when released, and it does. I believe that everyone who has seen the video agrees with that.

I personally also think that you can see the precession come to a stop over the space of 3 frames and not every one agrees with me for various reasons due to their own beliefs about gyro behavior. So be it. I always view what is happening in a gyro with simple mechanical motions, movements and forces, not magic, so I believe that motion cannot change instantaneously, it takes time to do so (I am not talking about forces but the magnitude of motion).

I can see where Ram (and my buddy Trent) is coming from for the third point about pivot reaction (although not everyone agrees with his view, but again, so be it). Ram's view is that IF the string is pulling on the tree house then the system is acting like the "rock on a string" scenario and therefore when the gyro is released it goes in one direction and the tree house goes in the other and IF this is the case then of course Ram is correct. Again not everyone thinks that this happens with gyros, so be it. We live, we learn some, and we move on, hopefully to learn some more.

Best to all,
Blaze



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Answer: Glenn Hawkins - 24/06/2012 18:29:17
 Hello Blaze, ‘ good to hear from you.

Blaze: “. . . I don't quite understand what you mean by the last line of the first paragraph.”

That line was, Glenn: “More tilting force must be added to overcome the resistance to increase a faster pace of tilting and all that goes with it.”

You Blaze continue, : “the problem is that there is always more than one of these motions and/or movements happening at the same time so we have to be aware of EVERY motion/movement that is happening to understand the total effect.”

Excuse me. I think you finally missed one. My explanation was not concerned with the total effects. Besides, I don’t believe in instantiations, action at a distance (magnetism), and same time actions. I believe in action, reaction, although multiple reactions can take place from one single action. (A cue ball hits two pool balls at the same time.)

When rotation is tilted there is a beginning. All else that happens consequently can be defined as “( and all that goes with it.)”. That is an all inclusive, blanket statement. The tilting, even minutely begins a series of reactions; tilt cause the reaction of deflection/resistance. Deflection/resistance causes two reactions, perhaps simultaneous which are torque upon the pivot and precession. I mean to say that if one ignores all else, and studies how and why tilt begins a chain reactions that force down on the pivot, he may come to understand why the wheel cannot rotate around it’s center of mass. I though that this would help, because seeing with the eyes is not always believing with the mind.

I will repeat, “More tilting force must be add to overcome the resistance to increase a faster pace of tilting and all that goes with it.” I want us to concentrate precisely on how and why torque forces down like an essence of weight on the pivot, until one may see in one’s mind, how and why the wheel must rotate/precess around the pivot, because the essence of the wheel’s weight is not under the wheel any more, but rather one may see that it has been transferred to the pivot. It is as if the wheel is setting on top of the pivot and therefore it is rotating around it’s center of weight, which is far removed from being under the actual wheel itself.

I have hopes you and your friends enjoy the day,
Glenn




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Answer: EDH - 14/08/2012 00:38:39
 Well, I haven't read every post in this thread, but I think that my answer is different from at least the original one given. If the spinning Gyro is attached to an arm which is attached to an unpowered freely rotating hub, and the Gyro is dropped under gravity, it will of course precess laterally and vertically nutate. If say, this Gyro is attached to it's arm by a small electromagnet and the switch is powered off, the Gyro will drop vertically to the ground with no horizontal movement whatsoever. If this same apparatus is horizontally powered by the hub, the Gyro will still precess at the correct frequencies,nutating up and down and generating strong pulsating vertical forces on the "hub". If this Gyro is again instantaneously disconnected, It will now move along a curved path in space which is a composite of both it's exit and precession vectors

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Answer: Blaze - 14/08/2012 02:06:15
 Actually, this has been settled. When the gyro is released it travels in a straight line until it hits the ground, just like a rock on a string. I did an experiment with a video that showed this quite well. The video link was posted earlier in this thread, however that link has expired so here is a new link with the same video. Please be patient when this loads as it does take quite a bit of time.

http://minus.com/lbnyZYnZOiPnNK

Blaze

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Answer: EDH - 14/08/2012 06:19:04
 Great experiment! Since a precessing Gyro has no angular momentum, does this mean that the ten inch displacement was reactionless?

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Answer: Blaze - 14/08/2012 17:01:03
 "Great experiment! Since a precessing Gyro has no angular momentum, does this mean that the ten inch displacement was reactionless?"

Please note that the actual measured movement was about 24 inches, even though the gentlemen in the video said it was about 10 inches. The lines on the white board on the grass were spaced 6 inches apart.

The answer to your question formed about of the responses in this thread. Some people believe the movement was reactionless, some people don't. It might be best for you to make up your own mind and leave it at that or otherwise risk another huge number of responses to this thread.

regards,
Blaze

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Answer: Glenn Hawkins - 16/08/2012 04:42:37
 Hello EDH,

I am delighted to see you. I hope you are well and happy. I’m going to call into question your post and disagree with you. Please know that I do not change my high regard for your splendid level of intelligence.

You say, “EDH - my answer is different from at least the original one given. If the spinning Gyro is attached to an arm which is attached to an unpowered freely rotating hub”

What is meant by ‘underpowered’?

You say,”. . . and the Gyro is dropped under gravity, it will of course precess laterally and vertically nutate.”

Do you mean, dropped while remaining attached to an arm, which itself is continuously attached to an unyielding string? Or, do you mean the wheel is unhinged from the string?

You say,” If say, this Gyro is attached to it's arm by a small electromagnet and the switch is powered off, the Gyro will drop vertically to the ground with no horizontal movement whatsoever.”

How can you say that when it is both, illogical for mass in motion not to continue in motion, and also after it has been shown in a seemingly undeniable experiment?

You say,” If this same apparatus is horizontally powered by the hub, the Gyro will still precess at the correct frequencies, nutating up and down and generating strong pulsating vertical forces on the "hub".

Nutation is rarely observed, because it requires certain usually avoided factors to be present. The idea of nutation complicate without warrant or value the understanding of this experiment. What good is to mention them?

You say, “If this Gyro is again instantaneously disconnected, It will now move along a curved path in space which is a composite of both it's exit and precession vectors.”

It is imposable for an object to curve in space without a constant deflection from a backdrop from which a force emanates to cause curving.

Answer: Blaze - 14/08/2012 02:06:15 Actually, this has been settled. When the gyro is released it travels in a straight line until it hits the ground, just like a rock on a string. I did an experiment with a video that showed this quite well. The video link was posted earlier in this thread, however that link has expired so here is a new link with the same video. Please be patient when this loads as it does take quite a bit of time.

http://minus.com/lbnyZYnZOiPnNK

Blaze

Answer: EDH - 14/08/2012 06:19:04 Great experiment! Since a precessing Gyro has no angular momentum, does this mean that the ten inch displacement was reactionless?

No that is an a fallacy entertained by a handful of people. There a the normal amount of angular momentum in play during precession. I knew it for 20 years, but it took Blazé to prove it. He too knew it before he designed and performed his test. What you see is exactly what how it should be, how it must me. It really is an excellent experiments. We can be prow of Blaze and his merry band of friends.



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