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Question

Asked by: Blaze
Subject: Calculating the "Drop" time
Question: The following is a simple first approximation of the Drop Time that occurs when an overhanging gyro starts precessing from a dead stop. This is a first approximation only. To accurately calculate the total drop time one would very likely have to integrate over time.

Consider a gyro system that has a precession rate, distance from pivot to centre of wheel (spinning mass), wheel spin, such that it gives a precession velocity at the wheel of 0.1 meters/second.

We know that the force or torque of precession is equal to the force or torque pulling the wheel down. Since these forces are equal we can assume that the rate of acceleration to steady state precession velocity is the same as the acceleration pulling the wheel down which is g or 9.81 meters/sec/sec. (f=ma, same mass Dropping as precessing, same force Dropping as precessing, therefore acceleration has to also be the same as well).

We also know that in the system described above we will have a steady state precession velocity of 0.1 meters/second. So to calculate the Drop Time we use velocity = acceleration x time or v=at, or t=v/a. Therefore for this system t=0.1/9.81 which is about 10 milliseconds or about 1/100 of a second.

It is interesting to note that the actual mass does not matter.

Any thoughts our there about this calculation?
Date: 28 April 2012
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Answers (Ordered by Date)


Answer: Luis Gonzalez - 28/04/2012 23:26:10
 Hi Blaze,
It is a rare pleasure to find another inquiring mind seeking similar things.
I tried this same set of operations some time ago.
And there is something even more interesting that results from these simple equation manipulations.

The rate of Acceleration also does not appear to affect the time (t) that the drop lasts.

The reason is that we are dividing the Velocity (v) of precession by the Acceleration (a):
t=v/a
And the Acceleration happens to be a factor in the Numerator of precession's velocity, therefore Acceleration (a) cancels out.
In my thinking, the implications of this possibility can be extraordinary.

Please tell me what you think Blaze.

Best Regards,
Luis G

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Answer: Blaze - 29/04/2012 00:28:19
 Interesting observation.

Let me start by stating why the mass does not matter. It is because it is the same mass. That is, during the drop, before steady state precession is achieved, the mass of the wheel is being accelerated down by gravity and at the same time being accelerated (for lack of a better word) forward in the plane of precession. It is the same mass that is being accelerated. It is just being accelerated in two directions at once. If you were to graph the resultant movement, it be moving at an angle, down and forward until full precession velocity is reached at which time no further downward movement would be detected (neglecting frictional effects of the pivot, of course).

The acceleration doesn't matter because it is the same for both directions, down for the drop and forward for the precession.

However, the velocity of steady state precession does matter. The faster the precession rate, the longer the drop time because it will take longer for a given acceleration rate to accelerate the mass to the steady state precession velocity. For example, if the steady state precession velocity (at the wheel centre) was 1 meter/sec then t=v/a would give a drop time of about 100 milliseconds or about 1/10 of a second.

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Answer: Glenn Hawkins - 29/04/2012 01:45:48
 I’m glad you posted this, because I am building a machine and there are mathematic like question to do with. . . why design for mass, shape and size.

You said, “We know that the force or torque of precession is equal to the force or torque pulling the wheel down.”

This cannot be true. A good gyroscope is apt to remain elevated for six munities, multiplied by milliseconds, but when the flywheel is crashed head-on, the momentum measured upon release into collision is perhaps one millisecond. The energy to support this six minute lift comes from the drop as the wheel slowly yields. Compared to that energy drained in time, the collision momentum is relatively week. As the flywheel is accelerated almost instantaneous “Momentous’ postulation” there is energy an drain too spontaneously quick to notice by eyesight. After that initial acceleration, precession acts in a near state of coasting and requires very little energy.

But Speed then, is what you infer also, the speed of drop verses the speed of precession. It is hard to argue with you there. You would seem to be correct and perhaps you are. But the magnitude of torque to constantly lift is a great deal greater that the torque to generate precession, more especially so because most of the movement in precession is due to coasting, not torque.

I hope you keep going with this. I don’t think a can continue. I have got to do stick-to-it-mess in my building effort and stop interring into everything that looks interesting. Good show. Good luck.
Sincerely Glenn,


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Answer: Glenn Hawkins - 29/04/2012 01:50:40
 Blase, your second post is right-on.

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Answer: Luis Gonzalez - 29/04/2012 02:50:49
 Blaze,
Your statements make sense; I agree with all of it up to now.
Your perspective is different but that's good because it reflects a different kind of thinking and that is welcome.
I also appreciate that your writing is clear with correct spelling and all.

How long have you been pursuing gyro dynamics?

Regards,
Luis G

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Answer: Blaze - 29/04/2012 03:21:14
 Thank you for your response Glenn. I have been silently reading these propulsion posts on and off over the years. Very interesting and I have learned a lot from you and the other “main players” who have posted.

I see things somewhat differently, NOT necessarily correctly, but differently. For the purpose of this discussion, let us talk about an “ideal” overhung gyroscope system. That is one where there is no friction of any kind in any part of the gyroscope system and the wheel is kept at a constant speed.

In this ideal gyroscope system, when the steady state precession velocity is reached after the initial drop, there should be no further movement down of the wheel. The precession would continue due to the force of gravity exerted on the wheel (the resultant of the down force of gravity is a force at 90 degrees, which is the precession force). It is this force of gravity exerted on the wheel that causes the continued precession. If you were to somehow magically remove the force of gravity, then the precession force would be removed also and the precession would stop. Think of a gimbaled gyro. You only need to apply a force (not a distance) to precess the gyro.

In a real world gyroscope system there are all kinds of friction involved so things change considerably. There is pivot friction, wheel bearing friction, windage losses, non spinning masses that cause unwanted movements and problems, etc. etc.

So in a real world gyroscope system the wheel continues to very slowly drop which is due primarily to the friction of the pivot. This friction has the same effect as putting a very, very small backwards force on the precessing gyroscope. As you know, if you put a force opposite to the direction of precession, the wheel drops in height. To make matters worse, as the wheel spin slows down, the precession speed increases. This means even greater friction from the pivot (mechanical friction generally increases with speed) and more backwards force.

So the down force (gravity) creates an equal resultant force that is 90 degrees to the down force. This is the precession force. Precession also creates an equal resultant force which is 90 degrees to the precession force. This is the up force which is what actually “holds up” the wheel in an overhung gyro system. The precession force is theoretically equal to the down force but in reality it is slightly less due to friction from the pivot. The up force is equal to the precession force, however the since the precession force is slightly less than the down force, the up force will also be slightly less than the down force and therefore the wheel slowly drops over time.


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Answer: Blaze - 29/04/2012 03:27:00
 Luis, as you can see from my previous post, I have been looking at this for a few years, but not with the intent of doing anything with it. It is just plain interesting. I like to work “outside the box”.

Just recently I came up with an idea that may have some merit but it is remarkably different than anything I have seen here or elsewhere. Doesn’t mean it will work, but I am checking it out anyway to find out for sure.


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Answer: Blaze - 29/04/2012 03:29:47
 By the way Luis, I have learned a lot from you as well.

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Answer: Luis Gonzalez - 29/04/2012 04:59:36
 LLLLLLLL
Thank you Blaze,
You have been doing your homework well.

If your idea has to do with inertial propulsion, then based on your expressed focus, my opinion is that you are correct and we are headed in the same direction.
If it is not about propulsion then I have no opinion.

The following item is something that you may have already considered too:
Since the time span (t) is independent from mass and independent from acceleration, then it is independent from force (f=ma).

That means that if we use the same exact device in any planet's gravity (of any magnitude) or we mechanically increase the force of the torque..., the drop time-span (t) remains constant (I think that is fascinating).
Of course this constant (t) is true only when all other factors such as gyro radius, spin velocity, torque radius, etc remain constant.
So each device when spun at same rate will always have the same (t), despite how much force is applied mechanically to the torque (within limits that will cause not cause anomalies). This is very important. It is as if each device has its own signature.

I look forward to your future postings.

Best Regards,
Luis G

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Answer: Blaze - 29/04/2012 06:35:25
 Sorry Luis, but on this one I have to at least partly disagree with you if I correctly understand what you mean. When I said acceleration doesn’t matter I meant for a given acceleration, like that of Earth which does not change and affects both the down force and the precession force on the wheel equally. That is why acceleration doesn’t matter, because it affects both forces equally and you don’t change it. It is the same reason the mass doesn’t matter, because you don’t change it so it affects both forces equally. If you change the acceleration (down force) then the precession speed also changes, which changes the drop time and drop distance. I guess I didn’t previously express myself as well as I could have.

If you take a gyro system on Earth that has a drop time of 1/63 seconds and put it on Mars (NASA says Mars gravity is 3.7 meter/sec/sec) with all other parameters kept the same as on Earth you will get a drop time of 1/167 seconds. This is because the reduced gravity will give a much slower precession speed (remember I said that the precession speed is important). My definition of precession speed is the speed that the center of the wheel is moving around the pivot. A slower precession speed means a quicker drop time because the mass does not have to accelerate to as high a precession speed and therefore it takes less time to get to the lower precession speed.

In order to keep the drop time the same on Mars as on Earth you would have to increase the speed of precession back to the same as that on Earth. The easiest way to do this would be to decrease the wheel spin to a value that will give you the same precession speed as when the system was running on Earth.

I have been using the terms precession speed and precession velocity interchangeably in these discussions even though technically they are not the same thing.

Blaze


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Answer: Blaze - 29/04/2012 15:29:12
 Perhaps another way to look at it is like this:

When using t=v/a, v is a known, not a variable , it is the precession speed for a given set of gyro system parameters. Those parameters include the acceleration or down force exerted on the gyro system. If you change the gyro system parameters (down force in this case), you will change the drop time as well because the precession speed has changed.

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Answer: Luis Gonzalez - 29/04/2012 15:53:51
 Hi Blaze,
Your logic is good but you have failed to consider one very important point.
Precession-velocity (v) is indeed a variable: Torque divided by the gyros momentum (you can look up this equation).

We know Mars's gravity has a Smaller magnitude, so the acceleration (a) is Smaller than Earths'.
We also know that the resulting precession velocity (v) is also slower/Smaller in Mars.

When we Divide precession-velocity (v) by gravity-acceleration (a) to obtain (t=v/a), BOTH (v) and (a) are proportionally Smaller quantities.
The RATIO happens to remain the same so (t) is constant, as long as all other gyro variables remain the same.

The reason (v) is proportional to (a) is that in all cases the Local gravity- acceleration is used in calculating the velocity of precession (precession v = T/L). And Gravity's acceleration (a) is a component factor of Torque (T).
So, when you divide (v) by (a), the (a) in the numerator, and the (a) in the denominator Cancel out.

If you need to go through the complete set of calculations to convince yourself that's ok.
Think through it and you will see the point. Then we can compare notes.

I believe this point can be of great importance for designing inventions.

Best Regards,
Luis G

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Answer: Blaze - 29/04/2012 16:23:24
 You are correct Luis. The drop time is the same for Earth and Mars. I have a spreadsheet setup where I can input the various parameters. I had (incorrectly) used a constant of 9.81 m/s/s calculations rather than using the input value for acceleration. I feel kinda sheepish right now.

So you are correct. Drop time does not change. However, drop distance does change. It is less for lower accelerations which makes sense as acceleration is used to calculate distance.

I guess one of the benefits about discussing things with others is that you get to “double check” your own ideas and calculations and find any errors one might have.

Thanks Luis.


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Answer: Luis Gonzalez - 29/04/2012 16:41:58
 Blaze,
That was quick work!
No one else has found a corrected path as quickly.

I need to do some gardening so I will be away for a good while.

Best Regards,
Luis G

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Answer: Harry K. - 29/04/2012 16:43:41
 Hello Blaze,

I like it how you see and explain gyro related behavior.
If you would use "torque" instead of "force" I would agree with most parts you have stated.
Led me describe how I see these things, which comply with most of your explanations:

An overhunging gyro produces a force caused by gravitation of the gyros dead mass with the distance R to the pivot point of the gyro and thus it is acting a torque. This torque increases if the dead weight mass increases or the radius (distance R) increases.

This torque causes additional angular momentum to the spinning flywheel and will be deflected by 90 degrees due to the degree of freedum of the overhunging spinning gyro system. The rsultant 90 degree defelected movement (Precession) is a compensation movement similar as a balanced level system, i.e. both sides, the torque of acting dead weight mass and on the other hand the precession movement are in balance.

All kind of friction forces but mainly the necessary angular momentum to accelerate the dead weigth mass to precession velocity (= rotation energy) as well as to accelerate the spinning gyro mass around its center of spinning mass in precession plane will cause a delay to balance the gyro system as described before. This delay causes the initial drop of the gyro, in theory as well as in reality. However, the angular momentum to accelerate the gyro to precession velocity ist stored and will cause in return an upward drop of the gyro if gravition would be suddenly removed for any reasons, because precession would running a small time after removment of gravitation until the stored angular momentum in precession plane would be absorbed.

To calculate the initial drop time you only have to calculate all counter torques caused by any kind of friction and the angular momentum to accelerate the dead weight mass of the gyro around its pivot as well as the angular momentum of the spinning mass around the center of spinning mass in precession plane.

To calculate the further continuous drop rate you have to calculated the sum of all possible counter torques caused by friction and then you can calculate the precession velocity for THIS sum of counter torque. With this rate of calculated precession velocity, the gyro will drop downwards. However, it is not considered that the tilting torque decreases during the drop movement of the gyro and that precession velocity increases if spinning velocity of the flywheel decreases by friction.
So it would be much more complex to calculate the drop rate over time in reality.

Best regards,
Harry




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Answer: Blaze - 29/04/2012 17:25:58
 Hi Harry. Thanks for your insight.

I think we are basically saying the same thing. I agree that there are a whole host of other factors to correctly calculate the actual initial drop and the continued drop. It may well not even be possible to calculate accurately, however, through experimentation one could establish a continued drop rate for a given gyro system. The simple formula t = v/a for calculating drop time will just get you “into the ball park”.

In my idealized gyro system with no friction of any kind, the system also would not have any non-spinning mass (which is impossible in the real world of course).

I agree 100% with you on what you call the “upward drop” (which I call “the rise”) that would happen if you suddenly removed gravity and the gyro precession comes to a halt. I call it the rise because it is exactly the opposite of the drop. The rise should theoretically be the same time and distance as the drop (neglecting all frictional effects).

Blaze


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Answer: Glenn Hawkins - 29/04/2012 22:55:14
 I think an ancient experiment I did belongs on this thread. I spun up to about 3,700 RPMs a Taco gyro and placed it on the plastic pivot. I hit the overhanging end of the shaft as hard as I could with a 20 oz. carpenter's hammer. The gyro held together every time. They are very tough. The light pedestal flew away in the air in one direction and the gyro accelerated in that instant to fling itself about fifteen feet across the room in the other direction. The experiment was repeated enough times that I believe the hammer surface landed a few times flat at a 180o. We all know about angles in collision and about a measure of resisting friction even between hard metal parts if the collision force is strong enough. So I leave the experiment to you to play with in your mind, rather than give you my guess work answers. The little experiment is significant I believe, and as indicated related to the subject here. It shows a great power transfer from drop-torque acceleration to precession.
Glenn,


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Answer: Momentus - 30/04/2012 11:13:39
 Id 1422 Gyroscope drop

Hello Blaze and welcome to the forum.

An elegant thought experiment, with some convincing mathematics. The fact that mass does not affect the calculation and that the time is constant is similar to the action of a pendulum.
With a pendulum, the means whereby vertical gravity can cause the weight to swing horizontally is obvious. In your post it is not mentioned??

The offset gyroscope has two motions.

First there is the gyroscope effect of spin, couple and torque, all at mutual right angles (orthogonal) which is well understood by classical science.

Secondly there is tangential velocity as the spinning disk orbits at the offset radius.

The gravity couple is as you say reacted by the gyroscope couple and precession, this occurs Instantaneously. The perfect gyroscope does not have time to drop. There is no vertical acceleration due to gravity. That is not disputed in classical mechanics.

This gyroscope action however does not create the tangential velocity.
By Newton’s first law there must be an external horizontal force acting on the disk. It is the absence of this force which constitutes the anomaly that shed dwellers seek to understand and exploit. I like Sandy’s “you can’t accelerate no mass” as a descriptive definition.

The drop which occurs with an imperfect system is due to the acceleration of “dead weight”, which cannot be instantaneous.


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Answer: Luis Gonzalez - 01/05/2012 02:14:21
 With all due respect to Momentus, I disagree.

No motion can accelerate from zero to a definite velocity (it is just impossible) and precession is no exception (even if it is just a change in the direction of an existing motion).

My opinion is that the drop and all the calculations mentioned in this thread apply to a "perfect" system.
Regards,
Luis G

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Answer: Blaze - 01/05/2012 03:21:23
 Hi Momentus. Thank you for your input.

You said “The perfect gyroscope does not have time to drop. There is no vertical acceleration due to gravity.”

That is not entirely accurate. Even a perfect gyroscope will have a Drop time and distance. This is because there are actually 3 forces involved but they are not equal during the Drop. That is why there IS a Drop. I am quite sure that you know about all three but some people don’t so I will go through them all for others who may be reading. The three forces are gravity (or down force), the precession force, and the up force.

The down force (gravity) creates an equal resultant force that is 90 degrees to the down force. This is the precession force. Precession also creates an equal resultant force which is 90 degrees to the precession force. This is the up force which is what actually “holds up” the wheel in an overhung gyro system.

At the “very start” of the Drop, the wheel starts moving down which causes a “tilt” in the wheel (one side is held up by the pivot, the other side is free to move). The result of this tilt is an equal resultant force that is 90 degrees to the down force. This is the precession force. However this precession force only has a maximum magnitude, it is not infinite. It is equal to the down force. The wheel has a certain amount of mass, so applying a limited force or acceleration to the mass of the wheel will require a certain amount of time to get to a certain speed (steady state precession speed in this case).

An instant after that “very start” of the Drop the gyro has started to precess but it is not up to full precession speed yet so its resultant force (the up force) is not equal to the down force, and so the wheel continues to fall. When the full precession speed is reached the up force is equal to the down force and the wheel no longer falls (neglecting all frictions of every kind, of course).

So, because there is a limited amount of acceleration or force available to bring the wheel up to steady state precession speed, it takes some time to do so. During the time it takes to get up to steady state precession speed, the wheel continues to fall at an ever decreasing rate because the down force acting on it is being offset by an ever increasing up force from the increasing precession speed.

I have been using the word force instead of torque but I think you know what I mean.

Blaze.


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Answer: Glenn Hawkins - 01/05/2012 04:20:28
 Momentous yours is one of the most elegant post yet. I think is beautiful and I like it and I heed it and learn from it. No it is not true, but sometimes I swear it seems almost true. A very small drop, a very short distance, can cause extreme speed of precession. (Remember my silly beating an overhung gyroscopes with a framing hammer?) The down distance and tilt from the hammer blow seemed fractional, but the precession (and yes I know a tinny bit of friction) caused the gyro to explode off the table with blinding speed of precession.

There is something else to consider relating to Blasé’s explanation. The condition of Nutations should be perhaps thought about. We see the wheel rise when precession jerks forward too fast. We see the gyro fall when gravity slows the rise. We see the precession speed increase when the gyro falls etc. It is a four stroke action.

The idea wheel would be a ring having something like an OD 10’x ID 9’ and it would have a shaft and fulcrum magnetically supported and accelerated like Professor Eric’s train. It would be tilted by magnetic repulsion and the collision force of precession would be captured by yet another magnetic repulsion system. Now then, the hole thing would be inside a vacuum. Will a set of such rings propelled us through the universe one day?

Very good Luis G. and doubly correct if that were possible.
Very good post Blasé and right you are.
Momentous, it was a very fine post, fader for the brain to helping me in design. Thanks to you I will now use shorter down strokes.
Glenn,


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Answer: Luis Gonzalez - 02/05/2012 04:00:45
 Hi Blaze,
There is one important point I need to clarify.
To me the term "Drop" is a generic reference to precession's acceleration from zero to its maximum steady velocity.
Inertia presents Resistance to precession's acceleration thus causing some colateral Downward motion (drop).

In other words, the downward motion is a resulting Effect, as precession's initiation encounters resistance.
And, the downward motion is NOT necessary to produce an initial impulse and drive precession.
Does that make sense?

Momentus writes in a concise manner and different style that always seems to elude my initial grasp. This has occurred previously, and it occurred here again.

I now see that Momentus is saying that the interaction between Torque and the initiation of Precession's acceleration does NOT need a Lag-Period requiring the flywheel to move downward under gravity's force (No Lag-Period means instantaneous).
Momentus is not claiming that precession's complete acceleration is instantaneous.

So, Torque ENGAGES with Precession instantaneously but requires time (t) to achieve its max steady velocity. The downward motion is not a necessary cause, it is a collateral effect from inertia's resistance.

Sorry Momentus, once again I failed to understand your concise word.
I don't think Momentus will respond, he usually doesn't.

Best Regards to All
Luis G

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Answer: Blaze - 03/05/2012 00:43:49
 You are correct Luis. Good clarification. I didn't mean to imply that there was a delay in the reaction of the down force and the precession force, only that, as you said, it takes time for the precession force to accelerate the wheel to steady state precession speed.

Blaze

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Answer: Luis Gonzalez - 03/05/2012 02:54:44
 Blaze,
Are you planning to build a device?
If so, when?
Regards,
Luis G

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Answer: Blaze - 03/05/2012 03:18:05
 Luis, I have one half put together and am waiting on parts to finish it but I doubt it will reveal anything new. More likely a new way of proving the same old thing.

Currently I am "building" a comprehensive explanation that hopefully reveals most or perhaps all of the hows and whys of a precessing gyroscope and why everything is not as it seems. My aim is to put it in simple enough terms that everyone can follow it. It will likely generate quite a bit of discussion. Don't hold your breath though as it will take a quite while to put it all together.

Blaze

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Answer: m - 05/05/2012 13:14:41
 

Quote “The three forces are gravity (or down force), the precession force, and the up force.”

No only one force, that of gravity. Pulling the mass down. As the far end of the shaft is supported, a couple is formed which rotates the system. That gravity couple is reacted instantaneously by the gyroscope. There is no build up, no delay, no drop.

Precession, the rotation of a gyroscope about its centre of mass also appears instantaneously. It is as wrong to say that precession causes the couple to form as it is to say that precession is caused by the applied torque. It does not happen like that. Precession and torque occur simultaneously.

There is no force generated by gyroscope precession. There is no Quote “precession force “. Forces come in pairs. If there were a horizontal force accelerating the mass of the gyroscope then there would be a reaction to that force at the suspension point.

That IS the mystery.

Having said all that, your comments do describe how the dead weight of an imperfect gyroscope are accelerated and why in real life a gyroscope does drop. An experiment which shows that the dead mass is the source of the drop is to add a small (relative to the gyroscope mass) dead weight, not at the gyroscope but on the other side of the support on an extension to the shaft. This reduces the gravity couple, hence the precession speed, yet as there is a greater mass to accelerate, there is a disproportionally large increase in the drop.

Yes the drop exists in experiment and you have described the sequence of events. This does not address the central mystery of why the mass moves at all.

Bear in mind if you would that what is known about the gyroscope is accurate. Use this to test your ideas as to the strange “Dark Motion” of the offset gyroscope



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Answer: Blaze - 06/05/2012 23:48:41
 I guess you could say my explanation was in more of “laymen terms”. It is not as technically precise as yours. Yes, gravity is the force acting downwards on the gyro, yes a couple is formed because the other end of the shaft is supported by the pivot and I never meant to imply that there was a delay in reaction to the gravity couple, the reaction is instant, agreed.

You are quite correct when you say forces come in pairs. Instead of saying “precession force” perhaps I should have said “reaction to the gravity couple is precession movement” and “reaction to precession movement is an upwards movement (or at least an attempted upwards movement) that balances the downward movement of the gravity couple”.

If there were no “upward force” (my words) or upward movement that balances the gravity couple, the gyro wheel would fall. I read somewhere a while ago that an easy way to “see” the resulting movement from an applied force like gravity is to simply rotate the direction of the applied force 90 degrees in the spin direction of the wheel. So, when you consider gravity in an overhung gyro, the force on the outside axle is down (the inside axle is being held up by the pivot which means outside axle has moved further than the inside axle). To find the resulting movement you would rotate the direction of the force 90 degrees in the spin direction of the wheel. So that would be a direction of movement of that axle (outside axle in this case) in the horizontal plain (clockwise or counter clockwise would depend on the spin direction of the wheel) and you can actually feel this when you move the axle of a spinning bicycle wheel. I think we both can agree with that.

Where it gets interesting is when you consider that the precession movement itself is causing the outside axle to be moving more than the inside axle (this time in the horizontal plain). Again applying the “rule of resulting movement” we would do a 90 degree rotation in the spin direction of the wheel which would give a direction of movement that is directly up. This up direction of movement doesn’t actually move the wheel up because it is being opposed by the down force, gravity (or the gravity couple if you like), unless the precession movement is “hurried along” in which case the up direction of movement overcomes the gravity couple and the wheel starts curving upwards. How would this upwards movement not be an “up couple” (directly opposite of a gravity couple)?

By the way, I get what the mystery is and I have conveniently left it out of my explanation above.

Blaze


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