Question |
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Blaze |
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Theory: Mass Transfer is really just Weight Transfer |
| Question: |
Premise:
What is commonly called Mass Transfer is actually ONLY weight transfer. The MASS of the gyro wheel stays with the wheel. Only the WEIGHT of the gyro wheel is transferred to the pivot.
Discussion:
Let us start with a definition of Mass Transfer. I will use the definition from a recent posting by Sandy Kidd which sums it up very well.
“The term Mass Transfer has been used by many of us for many years and simply relates to the fact that a gravity accelerated gyroscope system in precession has the ability to transfer all of its mass to act vertically down through the axis of system rotation, pivot point, fulcrum point etc. This is also true in many respects for mechanically accelerated systems.”
It is important to make a distinction between mass and weight.
The definition of mass is: “In physics, mass or more specifically inertial mass, can be defined as a quantitative measure of an object's resistance to the change of its speed.” “Mass is an intrinsic property of that body that never changes.”
So what this means is that the mass of an object is always there regardless of if the weight is there or not.
The definition of weight is: “Weight is the gravitational force acting on a given body, which differs depending on the gravitational pull of the opposing body” (a person's weight on Earth is more than a person’s weight on the Moon).
Think of an astronaut floating is gravity free space. His weight will be zero. His mass has not changed, just his weight. A more down to Earth example would be a person pushing a car. When you are pushing a car on flat smooth pavement you are moving the mass of the car. The weight of the car is being “held up” by the Earth. While you can move the mass of the car with muscle power you could never lift the weight of the car.
Theoretically if you were to put a weigh scale under the pivot of a precessing overhung gravity “powered” gyro you would measure the total weight of the entire gyro system. So the weight is clearly transferred to the pivot. However I believe that the mass stays with the precessing gyro wheel. I have previously theorized that the force of gravity being is being countered by an Up Force that is equal in magnitude to gravity and is what is holding the gyro wheel up (not everyone on this forum agrees with this theory). If this is the case then the gyro wheel is essentially “weightless” since gravity is being opposed by an equal force. Since the gyro is still in the Earth’s gravity field the weight has to be somewhere and it is transferred to the pivot (the up force is part of what acts to transfer the weight to the pivot). However that is just the weight, not the mass, much like the example of the astronaut or pushing the car.
So, how do we know that the mass stays with the gyro wheel? Well besides mass being an intrinsic property of that body that never changes, we know the mass stays with the gyro wheel because of the amount of force generated on whatever is stopping the wheel’s precession speed when crashing the wheel to a stop. The reason that “everyone’s” gyros appear to have very little or even no force when crashing the wheel to a stop is because the mass and speed of those gyro wheels are usually very small (this idea was postulated by Glenn Hawkins back on January 29 of 2008 and I believe he is correct).
Typical gyro wheels are usually a less than a pound, are spun up to some high rpm, have most of the weight concentrated around the outer rim of the wheel (this shape gives the slowest precession speed) and move at only a small fraction of a foot per second, so the force to stop this typical wheel would be extremely small.
For example, if you have a gyro system with following parameters:
Wheel mass: 0.25 kg or 0.55 pounds
Wheel diameter: 2 inches
Wheel rpm: 5000
Wheel design: most of the mass in the rim
Pivot to wheel distance: 2 inches
With this system the gyro wheel will have a precession speed of just under 3 inches per second or about ¼ foot per second or 0.075 meters per second. We can calculate the stopping force using F=ma and since a=v/t we get F=mv/t. If we assume a time of 1 second to stop the gyro wheel we get F=(0.25)(0.075)/1 which gives us a force of 0.01875 Newtons, very small. If we convert this to ounce-force we get 1/15 of an ounce for the stopping the gyro wheel’s precession speed. If you want to stop the mass in ½ second this would double the force to about 1/7 of an ounce of force. Even if you stop the mass in 1/100 of a second it would only generate a stopping force of about 6.7 ounces which is still small enough that most people wouldn’t notice it.
Look at it this way. If I were to “throw” you a 0.55 pound weight moving at 3 inches per second it wouldn’t throw you off balance at all. You wouldn’t even feel it. This would be more like gently handing you the 0.55 pound weight rather than throwing it. However if I threw a 55 pound weight at you moving at 30 feet per second you would be thrown backwards and probably land on your butt.
Now consider a large gyro system designed to maximize precession velocity with the following parameters:
Wheel mass: 50 kg or 110 pounds
Wheel diameter: 6 feet
Wheel rpm: 500
Wheel design: uniform thickness throughout
Pivot to wheel distance: 10 feet
With this system you will have a precession speed of about 13.7 feet per second or 4.16 meters per second and a stopping force of 46.8 pounds when stopped in one second, something you will be able to feel when you try to stop it by hand. If we stop this precession speed in 1/100 of a second we get a stopping force of 4,680 pounds, over two tons.
So, to get any significant stopping force you would need a really big gyro system to get the precession speed needed and a lot of mass to boot. Alternatively, one could apply additional down force to speed up precession but a large pivot to wheel distance is still required to keep the precession speed high and the precession period slow enough to prevent the unwanted slinging effect of the dead mass in the system.
The point of having a large stopping force is that it represents the amount of force or momentum that could be transferred to another object.
Your thoughts and comments?
Blaze
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| Date: |
24 May 2012
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Answers (Ordered by Date)
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| Answer: |
Glenn Hawkins - 25/05/2012 01:04:52
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| | Good post.
I am not sure at all the professor who coined the phrase meant how it has been interpreted.
GLENN'S MASS TRANSFER was always this.
A wheel has precessed from the start to stop of 1/2 orbit. Measure the diameter of the circle and that is the length mass has been moved without an opposite reaction. That is the only mass transfer I understand.
As to the explanation of the weight of wheel being torqued to the pivot, that's good.
Glenn,
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Glenn Hawkins - 25/05/2012 01:04:55
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| | Good post.
I am not sure at all the professor who coined the phrase meant how it has been interpreted.
GLENN'S MASS TRANSFER was always this.
A wheel has precessed from the start to stop of 1/2 orbit. Measure the diameter of the circle and that is the length mass has been moved without an opposite reaction. That is the only mass transfer I understand.
As to the explanation of the weight of wheel being torqued to the pivot, that's good.
Glenn,
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Blaze - 25/05/2012 03:58:02
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| | What you call GLENN'S MASS TRANSFER, I would mass movement (mass has moved without opposite reaction), not mass transfer. I would call mass transfer (or weight transfer) the "shifting" of the weight to the pivot. But that may be just me.
Blaze
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Ram Firestone - 25/05/2012 04:33:35
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| | Blaze Prof. Laithwaite coined the phrase "mass transfer" to mean moving mass without the full reaction. If you just say mass movement it might imply the full reaction. But this is all just mincing words. It doesn’t really matter what you call it as long as everyone knows what you are talking about. The bigger question is if it’s possible or not.
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Glenn Hawkins - 25/05/2012 14:59:28
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| | Very well explained Ram. We think in terms of words, Blase and it was good of you to see the need to address this.
Then it is mass movement for for me. Mass transfer for others. It is torque down on the pivot by gravity for me. In mechanically accelerated systems it is the same, except usually force is applied horizontally, therefore torque to hub. Nobody needs to believe anything, but we have quick names with distention and the explanations for those who do not know about conditions of weight , mass and other things can be made to give understanding to the names.
My mind is better set and I can use these words to think by. To each his own. Thank you.
MASS MOVEMENT
MASS TRANSFER
TORQUE DOWN
TORQUE TO HUB
Glenn,
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Blaze - 25/05/2012 18:20:54
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| | So you guys agree that there would be significant stopping force required for a large dimension, heavy mass gyro system? This is important as it is leading up to the next step/theory.
Blaze
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Sandy Kidd - 25/05/2012 22:25:35
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| | Hello Blaze,
If we were discussing accepted physics I would tend to agree with your statements but I always tended to follow these rules, and I will now put the cat among the pigeons.
Any object not under acceleration or braking does nothing.
Any object in space not subject to acceleration or braking may just as well not exist.
So without acceleration or some kind of gravity we have no mass, which is the other way round.
Does it matter if a space rock has “x” million bits of this, and “y” million bits of that, which go together to make up its mass, without acceleration it just does not matter.
So if the mass of the gyroscope is still present, and the gyroscope is still in rotation or precession, then why is there no angular momentum?
If the mass is still there and the system is in rotation where has the centrifugal force gone?
It would appear that there is no accelerated mass in the system.
There is nothing to accelerate.
About this time I coined the phrase “You cannot accelerate no mass”
This phrase was coined many years ago and in fact was the title of my first or second posting to this forum in 2004
At this time I was testing twin opposed offset gyroscopic systems and many of those were being mechanically accelerated up to 1000rpm.
When the gyro rotation speed was of a sufficient order to create what some people call precession (I later called this saturation in an effort to avoid confusion) it was discovered that like gravity accelerated systems there was effectively no mass left in the system.
Any increase in system rotation speed I thought at the time would increase the acceleration rate of the system and return the gyros to a position around the horizontal.
But gyroscopes being gyroscopes this was unfortunately a bad guess.
I went to some quite elaborate lengths to prove I was not mistaken with what appeared to be going on.
I was not mistaken.
There was no mass left to accelerate.
So I ended up watching a pair of massless discs being rotated at high speed without any acceleration.
I will not try to tell you how many times I repeated this experiment because I just could believe I was watching the impossible.
This phenomenon is unique to gyroscopes and physics
Also in gravity accelerated systems the gyroscope rotates around its centre of no mass.
In the final analysis and in this context mass only exists because you can see it.
Sandy Kidd
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Blaze - 25/05/2012 23:40:05
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| | Thanks for your comments Sandy. I think I follow you. I agree that if something is just floating free in space it affects nothing, unless you try to move it. Now IF (and that is a mighty big IF) I understand the mechanically accelerated system you described in this and other much older posts, then I would think (maybe incorrectly) that the reason the wheels rise is for the same reason the wheel rises in a gravity powered gyro that is pushed faster than its normal precession speed. The “up force” (my term) is now larger than the down force (gravity) so the wheels rise. As to why there is no angular momentum or centrifugal force when precession is sped up to beyond the normal speed, I don’t know.
The BIG problem is that if the wheel no longer has any mass at normal precession speed, and then the axle to the gyro wheel snapped (like a string holding the rock swung around your hand) the gyro wheel would just STOP. I would not keep moving in a straight line. This is something I have been debating with a few people about at work. Their theory goes as such. If there is no mass at the wheel at the time of release, then momentum = mass x velocity = zero because the mass is zero. So, no momentum means no movement. Personally, I have a hard time believing that a wheel with a precession speed of, say 10 feet per second, could just suddenly stop if it were suddenly disconnected from the gyro arm. To me that just doesn’t make any sense, but then, little does with overhung gyros.
As a result of those debates and a seeming lack of really large gyros, I have started building a large gyro using a 16 inch car rim and tire with a 4 foot distance from the pivot to the wheel. This system should provide a stopping force of about 13 pounds when stopped over the time of one second, which should be enough to settle the question of whether there is any mass or momentum at the wheel. Of course I realize that momentum is only felt when the precession speed is actually changing which is exactly what I am trying to do, but if, in fact, there is a significant stopping force for this system, then the wheel should move off in a straight line tangent to the precession circle if the wheel were suddenly disconnected from the pivot arm. At least that is my theory.
Wish me luck,
Blaze
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Luis Gonzalez - 26/05/2012 16:55:36
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| | Hi Blaze,
Your insight to parse the terms “Mass” vs. “Weight” into correct usage, bring out to the open-light some concepts that have been buzzing around in in the semi-dark twilight of this forum for some time now.
Well done.
Your ideas are definitely refreshing toward clearer understanding of the “displacement” notions discussed in this forum.
I Also hope I am not alone in understanding that the displacement of “Weight” (not mass) “to the pivot point” of a “traditional” gravity-driven gyro, is intimately related to the "Reactionless" displacement (or movement) of the flywheel’s “Mass” along the curve of an arc. Without one you would not have the other.
How the different terms have been previously used (or misused) in this forum is LESS important than what concepts and interactions bring clarity to the observed and perceived phenomena.
I personally prefer the terms you use, they remove some of the effort needed to fully grasp and explain this convoluted subject.
Best Regards,
Luis G
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Luis Gonzalez - 27/05/2012 05:18:26
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| | Hi Blaze,
I am again enjoying your challenging question in the above posting.
The concept behind this question has emerged in this forum a number of times in different sets of clothing.
I like the way you posed the question …if the gyro axel snapped, will the gyro-wheel stop, or keep moving in a straight line?
This question paints a mental picture and offers a choice between two possible answers; it poses a mental experiment, which is a powerful tool.
My initial thought is that IF there is a drop-down (when the gyro revs up to steady state), then some degree of precession momentum will continue along the straight-line.
The connection between these 2 events is “momentum”.
The drop-down occurs as precession “accelerates” over a period of time, which indicates that mmentum is accumulated.
If accumulation of velocity/momentun were Not necessary, then we could be sure that precession’s motion does not act to store some degree of “momentum”.
Regards,
Luis G
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Blaze - 27/05/2012 05:37:03
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| | Very good Luis. I have been thinking for some time that an overhung gyro might be a momentum storage device, much like a capacitor stores electric charge.
I agree with you that the flywheel builds up momentum during the drop as it accelerates to normal precession speed. Once precessing, if one could turn off gravity the flywheel would then rise as it decelerates using up its store of momentum, the driving force causing the rise.
Blaze
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Luis Gonzalez - 27/05/2012 16:38:45
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| | Hi Blaze,
Agreed.
Regards,
Luis G
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Luis Gonzalez - 28/05/2012 23:18:06
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| | Hi Blaze,
We know an orbiting satellite requites gravity’s Centripetal “PULL” to keep the satellite (object) from continuing in straight-line motion (momentum).
Most of us also know that gyro-precession originates at the flywheel’s center of mass (inner precession).
This Flywheel force also drives its mass to orbit around the External pivot point (i.e. external to the flywheel’s center of mass) by “PUSHING angularly” against the remote pivot, NOT by being Pulled from the pivot at a distance (in a previous postings I refer to this angular push action, as “Unfolding”).
The mass of the spinning gyro flywheel is kept from continuing in a straight line by an angular “PUSH” provided by the dynamics within the spinning flywheel itself.
This “Push” provides the necessary “Centripetal PUSH” (not Pull) to keep the flywheel revolving around the pivot (perhaps with a little help from friction).
At steady-state, (a) the flywheel’s mass provides the straight line momentum i.e. “Centrifugal”, and (b) the spinning flywheel’s dynamics provide the countering Centripetal PUSH-IN, rather than being pulled into the pivot (somewhat different from the revolving satellite).
(Is Centripetal mixed with some pull…? Perhaps!)
The absence of Pull (or reduced pull) from the pivot-center is the reason that precession’s orbit shows reduced centrifugal force.
Note 1 - This explanation applies primarily to gravity driven gyros.
Note 2 - We should not confuse the above explanation with a yet different explanation regarding why Sandy’s experimental System (using a mechanically-driven Rotating-System) appears to lose centrifugal capability after the gyros are spun-up.
Best Regards,
Luis G
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