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16 June 2019 04:16

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Subject: Challenging Questions
Question: I am going to post scenarios with challenging questions. From this I hope to find knowledge from community reasoning. If there will be any response I will add to as we go along.

Two equal flywheels in space are set at opposite ends of a shaft, just like the rear wheels of a car and both are spun in reverse at equal speeds. One side is caused to move downward in a curve shaped like an arch. (as if dropped, if you will) The other end is locked such that it cannot move up, or down. (as if setting on a pedestal, if you will)

It might seem surprising that both would attempt to precess from an equal force, but since precession force is owing to the tilting degrees, not the distance of the drop, the total initiating force is the same. The distance and speed of procession is greater at the ‘drop’ wheel, but the torque in much greater at the (pivoting) wheel. This order and distribution of force is exactly the same as the principle of the lever.

These conditions can be duplicated in space with mechanically arranged mechanisms; to what results?

Both are precessing forward with equal 'over-all' force; the ‘dropped’ wheel precessing in a large arch and the ‘pedestal’ wheel precessing in a small arch.

Would the ‘pedestal’ end of the shaft then by its own power be kept from reacting in an opposite direction from that of the ‘dropped’ wheel? Your thoughts?

Incidentally, we see perhaps the same happening when an ice coated pedestal is sliding around a slick surface. The pedestal follows the wheel as it curves, but in a smaller arch than the wheel and all action in this way is uniform and constant.

Glenn,
Date: 11 June 2014
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Hi Glenn. If you ignore what happens during the drop and the drop distance, the two gyros will barycenter around a point at the middle of the shaft. Even if you don't ignore what happens during the drop and the drop distance, the result will be the same but the plain of barycenter precession will be slightly different. This is assuming that whatever is preventing the pivot gyro from moving up or down has no mass and does allow the pivot gyro to move freely horizontally. This also assumes that I understand the question correctly.

cheers,
Blaze

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Answer: Glenn Hawkins - 12/06/2014 03:28:32
Thank you Blase,
I knew you'd come through. Reconsider that both side-by-side flywheels separated by a shaft are rotating in the same direction; both tops north, bottoms south. They are tilting in the same direction and attempting to precess in the same direction; one in a small circle; the outer in a large circle.

I can try another explanation if this doesn't get it. I think this is an interesting puzzle in attempting to understand what I.P. function might work, if anything might work.

Cheers,
Glenn

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Hi Glenn. If you are talking about a system in space where there is no gravity and the input force (which would normally be gravity) is provided by some mechanical means, then there is no reason for the pedestal gyro to move in a small circle except if
1) it is either forced to do so by some outside means or
2) the mass of the pedestal itself is large enough that the barycenter is close to the pedestal gyro.

Other than that the system will barycenter around the middle of the shaft. Again, this is assuming that whatever is preventing the pivot gyro from moving up or down has no mass and allows the pivot gyro to move freely horizontally.

regards,
Blaze

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I have also assumed that the pedestal is directly underneath the pedestal gyro. If it is offset, then other conditions may apply.

cheers,
Blaze

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Answer: Glenn Hawkins - 13/06/2014 03:36:48
Blaze, I am attempting to express the same question in each of the different images I have offered so as to try to make the mechanics understood. They can't be answered until the are first understood. It is always so difficult, as my mind has obviously devised different ways of thinking than usual, but the subject is difficult at any rate. Try to delay thinking about the barycenter until later.

Try this. A dumbbell is rotating. One blob is magnetized and levitates a fraction of an inch above a repulsing opposite pole magnetic table. The other (unmagnified) blob is dropped from a 45 o . What will the blobs attempt to do?

It is a convoluted question and not so simple as it appears. I will try another explanation of the question later.

In appreciation,
Glenn

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Hi Glenn. I think I got it now. If the pivot gyro (high rpm gyro) were not allowed to move horizontally, then the two precession speeds from the high rpm and low rpm gyros would end up as one precession speed that is a compromise of the speed that the two gyros want to precess at. The drop distance during the time coming up to precession speed would also be larger.

cheers,
Blaze

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Answer: Glenn Hawkins - 14/06/2014 23:39:46
Very good. But sorry; the levitating wheel may move horizontally, but not vertically.

Each time that I was not understood I would try to make the question simpler. So before we get back to the advanced conditions concerning different rotating speeds, let us know that for now the shaft and wheels are one solid part, as a dumbbell and that everything, shaft and wheels spin at the same speed. We are getting there.
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The dropping wheel will attempt to precess around the levitating wheel.
The levitating wheel will attempt to precess around the dropping wheel.
Which wheel will be dominant and overpower the precession of the other?

Consider that the magnitude of precession force is not determined by the distance of the drop; but by the degree of vertical tilting each wheel performs and that these two perform the same degree of tilting. We should delay considering the different speed and power of torque the wheels are under. Next time maybe. Baby steps for me; one at a time.

Only the length of the shaft will determine a veritable difference as to which wheel overpowers the other and by how much. This is because the shaft is acting in the principle of the lever. The greater the distance of the dropping wheel, the greater the force ratio applied against the levitating wheel.

What will this cause? Increasing the length of the shaft will cause the barycenter to move toward the levitating wheel, which would be in conflict with center point of gravity thorium whenever the wheels are not rotating.

Here and we might not have thought about it, but torque generated can overpower the natural barycenter of gravity.

Something else important comes in to play, which is 'centrifuge' and because of that something extraordinary begins to happen. I am getting ahead of myself.

Also later we can consider what will happen if the wheels are rotating oppositely and other things, but this is a plate full enough for now.

Thanks,
Glenn

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Hi Glenn. For your scenario with the flywheels welded to the shaft and everything spinning at the same speed, the response is very simple. You said the "pivot" flywheel can move horizontally so I will assume that means horizontally in any horizontal direction.

The response is this. The two wheels will barycenter about the mid point of the shaft. Here is why.

Each flywheel provides a twisting torque to get up to precession speed. That torque is applied against the opposite wheel via the connecting shaft. The torques are equal because the wheels are identical and spinning at identical speeds. Therefore NEITHER wheel will be dominant in its response. Each wheel is pushing against the other (via the connecting shaft) to bring itself up to the exact same precession speed. So the system will barycenter around the mid point of the shaft, neglecting any friction in the system.

regards,
Blaze

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One more thing. The drop distance for your two flywheel system will be 1/2 of the normal drop distance of a typical one flywheel system (where the flywheel is not at the pivot end).

cheers,
Blaze

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Answer: Glenn Hawkins - 15/06/2014 23:32:31
Yes Blaze, your response is simple in following classical physics and your own experimentation, yet I am sure am sure you are wrong. I started replying, but I am exasperated with this communication already as I became with all others efforts before it and here I have hardly even scratched the surface of the beginning. There have been fifteen posts of mostly repetition to get to here and there was so many more to go. I think I will apply my mind and time elsewhere. If I should ever be compelled to explain something again, it will be with show and tell as has been done better by those before me. (I.E. Video) Thank you for your thoughtful input. You are a sharp guy and appreciated and what you believed is believed by almost everyone. ‘So long. I have to get something done. and this isn't it.
regards and tribute,
Glenn

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Answer: Glenn Hawkins - 16/06/2014 03:02:58
On second thought: I will skip the foreplay of building to understand and get to the orgasm.

Form a clear plastic disk in the shape of an inflated auto inner tube. Connect it somehow to a shaft through the center of this see-through plastic disk and set one end of the shaft on a pedestal and the other held ready to drop from a 45o angle.

First however, insert spongy muff balls free to move, then rotate the disk and drop it to curve downward. The balls will be under the influence of three directional forces.

Centripetal, the balls will be pressed uniformly outward toward rotation.

Inertial resistance, the balls will be pressed sideways at 180 degrees from centrifuge as they resist the direction of tilting. This is constant as centrifuge keeps attempting to realign the balls outward.

The sides of the balls are crushed inward at 90 degrees from one another as they rotate inside their concave enclosure. If the right angle crushing resistance from tilting would become stronger by increasing the tilt speed, the balls will move further away from the centripetal direction, which allows the inner two-way crushing forces to now converge and apply against the concave disk at a dual inside 45 degrees direction.

Centrifuge can have great momentum and it is now applying the crushing of the balls at a 45o angle inside the concave enclosure. The resistance to tilting is also being pulled to the area toward where the balls once were before they were tilted. Therefore both force and resistance apply against the 45o angle inside the concave enclosure.

The ultimate force direction for the disk will be in the force shape of a cone. The cone will be sideways with the nose toward the direction of tilt, but with the outward force toward the base of the cone, which is the opposite direction to tilt and spread.

From here you only have to waggle the top, or bottom of the disk, but not both at the same time, in order to cause the wheel to pull itself in a wider circumference. As it pulls itself, the amount of pulling force how ever much it is does not apply a rearward equal and opposite reaction against the pivot.

I have done this in experimentation, but friction and the difficulty of causing high acceleration into tilt without even greater friction are such that my reading are not trustworthy. But in this way I believe I have successfully created linear thrust from rotation.

If no one understands my excellent expository writing skills so be it, for I know I have done it well leaving no need for ifs and buts responses to my words; though perhaps to the mechanics. I purposely left out the complicated play of sideways momentum; and yet I have accomplished what I set out to do. Within this communication for those who can see it, is the explanation that a barycenter does move micro millimeters invisibly in the diminutiveness of distance toward the pivot with the application of even a small torque.

Goodbye for a while. My argument is finished and I really have to stop and do work. This is draining and unproductive and damn well not simple, but it has been a sometimes pleasure conversing.
Glenn

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Answer: Patrick Hill - 18/06/2014 02:45:27
Inflate the muff balls of course,but let's just say anyone put pen to old paper to me could havecorrespondence with a missed space,of photo copies of my work sent back,overveiwing forum antics I chuckle....as spoken word from myself becomes distorted so let's see who writes to me first,to receive the first and all ways conveyance..... ...........................what if primes ........about 360.....Egyptian math.....knights of templar.........and silly me son of a wildbore.........uncle one of heads of sellafeild nuclear power plant...other uncle house of lords for (defra) . ..le2 8eg.43.

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