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Question
Asked by: Harry K.
Subject: Is there a balance point and may it cause thrust?
Question: Hello all!

Is there an individual balance point where centrifugal and precessional forces of a gyro system are balanced? (Gravitation and inert forces must also be considered!).
May it be possible to cause thrust with the help of this balance point?

I would like to introduce here my theory of balance point between centrifugal and precessional forces of a gyro system. This is only an idea of an incomplete and may be a wrong theory.
This theory is simple, may be too simple! Therefore I have my doubts about its rightness.
The idea was born when EDH described his idea of a gyroscope propulsion system explained in this thread:
http://www.gyroscopes.org/forum/questions.asp?id=887

I will also present sketches and all necessary equations for calculations. Any comments would be appreciated!

Regards,
Harry
(Harald Keipert)
Date: 27 May 2008
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Answers (Ordered by Date)
Answer: Luis Gonzalez - 27/05/2008 13:51:16
 Dear Harry,
I look forward to this.
Best Regards,
Luis

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Answer: Glenn Hawkins - 27/05/2008 18:03:54
 sketches :))! Oh yeah! ehawkins32@comcast.net Thanks,

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Answer: Harry K. - 27/05/2008 19:11:24
 Hello all,

Okay, let's begin. I would like to make a step-by-step explanation to avoid misunderstandings. So please forgive me if you feel that my explanations are too common for you.

1. Experimental setup

I have uploaded a sketch on my server which shows the experimental setup for better understanding. Here is the link of the PDF-file (please copy and paste the link in your broser's URL address box):

http://www.misc.keipert.net/gyro/Gyro Forces 1.pdf

The sketch shows the front view of the gyro setup. On the left side there is a vertically aligned gyroscope. For understanding it is not important whether the spinning shape is rim, disc, sphere ore anything else. I have shown it as a cross section of a rim. Unfortunately I can only make drawings in 2D but not in 3D. I hope you can imagine it nevertheless?

This gyro axis is rigidly coupled to a lever. This lever is installed to a fulcrum at the center of hub in a way, that the gyro with the lever can freely pivot in vertical plane around the center of hub (presented by the double-headed arrow on the sketch).

The gyro with the lever can be freely rotated around the center of hub by a drive in horizontal plane.
The gyro itself can be freely rotated around the center of spinning mass in 90 degrees direction of the momentary lever axis.
Both rotaion possibilities are not linked together, i.e. they are independent from each other.
The gyro system is rigidly fixed onto the ground with the hub.


2. Opertion conditions

Please note that gravitation and inertial forces for moving masses around the pivot of the hub are not considered at the moment! I will come back to these (also important) forces later.

- The gyro is spinning with a defined velocity and in CCW direction (from side view) of rotation.
- The spinning gyro is rotated around the hub by a defined velocity and in direction (from top view) of rotation. This rotation is a forced precession of the spinning gyro system.

The following torques and forces are now acting at the gyro system:
1. Precession torque Fp * lp
2. Centrifugal force Fz
3. Centripetal force -Fz

The precession torque is equal to the torque, caused by forced precession with the drive at the hub. The resulting movement of this configuration will be an upward rotation of gyro and lever in vertical plane around the center of hub, i.e. the gyro will arise. Centrifugal and centripetal forces have here no effect at all, because the direction of these forces are congruent with the axis of rotation axis of the spinning gyro.

If we do not consider a further influence of centrifugal and centripetal forcees, the gyro with its lever would end its upward movement at he vertical top position (90 deg. to horizontal plane) and remain at this position, because the spinning plane of the gyro is here parallel to the rotation plane of the hub.
The tilting torque, caused by forced precession will decrease from maximum at horizontal position to zero at 90 deg. vertical position. This decrease of tliting torque is processed by a sinus function.

I will make a break here and continue later or tomorrow. I hope all is comprehensible so far? If not please let me know. Thanks!




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Answer: Luis Gonzalez - 28/05/2008 01:17:36
 Keep going Harry,
I am curious to see how your theory develops.
Best Regards,
Luis


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Answer: Glenn Hawkins - 28/05/2008 13:22:25
 To this point, this is a good enough description of the most basic actions of the gyroscopic. It is doubtlessly a necessary beginning for the assortment of laypersons that may visit.

Since our new beginning is to help each other clarify to the simplest presentation possible I offer the following.

You might put the minus sign in for FP in the drawing to the right. Add CCW to the drawing. Your pivot point is elevated above the hub? Every line is straight and the drawing is box-like. There are no angles, or curves.

Please expound on: “Centrifugal and centripetal forces have here no effect at all, because the direction of these forces are congruent with the axis of rotation axis of the spinning gyro.”

If I were sure I understood I would disagree.

So far it is like the man who fell from a ten-story building. As he passed each floor he shouted through the windows, ‘So far, so good.’ Thank you Harry. As you go along I look forward to further understanding what your fine mind has envisioned.

Regards,
Glenn


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Answer: Harry K. - 28/05/2008 15:51:44
 Hello Glenn,

Thank you for your feedback! I know the drawings are abstract and maybe not easy to understand, but as already mentioned I only can provide 2D-drawings.

I have updated the drawing and implemented the dirction of rotation. Please note, that the forces are vectors and thherefore the arrow shows the direction of the force. A "minus" value of the force would state, that this force is acting in counter direction of the arrow. I hope you understand.

Regarding my statement "Centrifugal and centripetal forces have here no effect at all, ..." I have to admit that this definition is only correct for the described operation condition:
1. The gyro is spinning first.
2. After the gyro has accelerated to the defined rotation, the gyro system will be rotated around the hub.
At the very beginning of rotating the the gyro around the hub, centrifugal forces would have very less influence, because the precession torque is much more dominant.

Sorry if I have caused confusion.

Best regards,
Harry

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Answer: Glenn Hawkins - 28/05/2008 16:31:47
 Dear Harry,

It is clear and accepted.

Regards, Glenn


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Answer: Harry K. - 28/05/2008 17:16:41
 Hello all,

Sorry for the long break but I had to prepare some more drawings.

2. Opertion conditions (continuation)

Now I will change the operation conditions in that way:

- The non-spinning gyro is rotated around the hub by a defined velocity and in CCW direction (from top view) of rotation. This rotation is equal to single dead weight mass, which is rotating around the hub. Therefore centripetal and centrifugal forces are acting.
- The already rotated gyro begins to spin to a defined velocity and in CCW direction (from side view) of rotation. Now there is acting a combination of centrifugal and precessional forces on the gyro system.

Here is the link of the second PDF-file (please copy and paste the link in your broser's URL address box):

http://www.misc.keipert.net/gyro/Gyro Forces 2.pdf

Now thee is acting a resultant force FR, which is caused by precessional force Fp and centrifugal force Fp. The resultant force Fr causes a torque around the pivot with lever arm lr in CW-direction:
TR = FR * lr

This torque will cause the lift of the gyro in same CW-dirction as the precessional torque Tp. This behavior is not new but the fact, that the torque TR may be higher than the precessional torque Fp, depending on all given parameters!

One note to the text "counter force to keep system in balance" in the drawing:
This force must be applied (e.g. by the ground) to avoid that the hub with the gyro does not move away in direction of vector FR. This is only necessary because there is no second counter gyro present for stabilisation.

3. Gyro rising above horizonntal plane

The next sketch shows the gyro in a momentary angle above horizontal plane.
Here is the link of the third PDF-file (please copy and paste the link in your broser's URL address box):

http://www.misc.keipert.net/gyro/Gyro Forces 3.pdf

As expected before, the gyro has arised to a certain angle after a certain time. The only difference to the second picture is the fact, that the centriptal force will now cause an additional torque around the pivot with lever arm lz in CCW direction!
However the resultant torque TR = FR * lr still causes a movement of the gyro system in CW-direction and therefore the gyro system still arises, but decelerated.
Please also note, that forces Fz and Fp are now similarly smaller, because the inward rotation causes a smaller radius of rotation around the hub.

4. Reaching the balance point

The next sketch shows at last the balanced gyro.
Here is the link of the fourth PDF-file (please copy and paste the link in your broser's URL address box):

http://www.misc.keipert.net/gyro/Gyro Forces 4.pdf

The difference to picture number 4 is essential !. The resultant force FR is now acting with the same angle as the spinning axis of the gyro. That means, that FR will not cause a torque again, because the distance lr to the pivot is zero.
In this scenarion, all torques in CW and CCW direction are canceled by Fz * lz = Fp * lp.

There is only left the resultant force FR, which is still acting at the pivot with the angle "a".

5. Cause of Thrust (?)

The last sketch shows an assemble of two balanced gyros, mounted in opposite direction.
Here is the link of the fifth PDF-file (please copy and paste the link in your broser's URL address box):

http://www.misc.keipert.net/gyro/Gyro Forces 5.pdf

Because the sum of all acting torques is zero in a balanced system, only the resultant force FR is acting via the lever at the pivot. To avoid a movement of the gyro system, an additional counter force -FR must be implemented.

Now another gyro will be assembled at the pivot at the opposite direction. Two forces FR are now acting upwards under the angle "a".
The picture on the bottom of the sketch shows again the final resultant force FT, caused by the two resultant forces FR.
This final resultant force FT is acting in vertical direction and should cause thrust.

Now you can lough about me and my theory, however, it would be nice, if you could explain clearly where my errors are. I have already mentioned, that this theory seems to be too simple so that there should be one or more mistakes in understanding.

If you do not find the mistake(s) at first go, I will continue and present the necessary calculations.

Thanks!
Harry




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Answer: Glenn Hawkins - 29/05/2008 16:04:03
 Dear Harry,
Don’t feel forlorn. I’m working in spurts. Energy level is up, but demands on my time are suddenly, temporarily high. Be with you soon.
Regards, Glenn


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Answer: Glenn Hawkins - 29/05/2008 20:34:27
 Dear Harry,

There is no flaw in your theory, but it is limited. The acceleration is only one to two feet upwards depending on the size of the apparatus. I am amassed even so. So congratulations you are right! I’m happy for you.

For all those visiting this thread go directly to http://www.misc.keipert.net/gyro/Gyro Forces 5.pdf and you will immediately understand Harry.

The upward arching of the gyro is continuous motion, until it reaches top dead center. There is no balance point. It isn’t possible to increase centrifuge without also increasing the force of the deflections that create lift. Lift is a continuous action first explained by our eldest member. Your drawing is a stop frame action of motion to explain what is happening.

It suddenly dawned on me what EDH is not saying. I now reverse myself and predict that his device will work. It is totally different than what I’ve done, yet the series of force reactions to achieve the gold I and at least two others (Hi Sandy) have been championing for a long time.

I will not divulge what EDH has done. That’s his secrete.

Your mind defies physics, Harry, but it is correct. What do you think about that?

Regards,
Glenn


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Answer: Glenn Hawkins - 29/05/2008 23:18:52
 Hi Harry and everyone,

This is very strange and complicated. There’s more to it than expressed. I will finish it if I’m capable. I Don’t know when that will be. Now I have completely forgotten what I found in Sand’s post so important. Also the computer ate my notes. I’m hunting again in his post for the information that jarred me and expanded my mind. I’ll try to look at this again when I can. I’m concerned with that old hissing adder that won’t go away, the pivot reaction in the precession plane.

My Regards,
Glenn


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Answer: Luis Gonzalez - 30/05/2008 01:50:47
 Hi Harry,
Here are some comments about your initial posting:

You said:
“- The spinning gyro is rotated around the hub by a defined velocity and in direction (from top view) of rotation. This rotation is a forced precession of the spinning gyro system.” (you missed ‘CCW’ in this sentence.)
*In the above sentence you are using the name “forced precession” in reference to the rotation around the hub from which the input torque is derived.
This creates ambiguity regarding what we should associate with the word “precession”.
I would overlook it now, but I think it creates room for miscommunication in future dialog when we compare the magnitude in the force of the input-torque Vs the magnitude in the force of the precession-deflection.
The input force should never be referred to as precession or any word derived from precession.
We can say that the rotation is the cause of precession or causes precession (keeping good cause & effect form).
I am sure you understand what I am saying and please let me know if I misunderstood your sentence.

You Said:
“The precession torque is equal to the torque, caused by forced precession with the drive at the hub.”
*I believe this is the same misnaming issue that I mentioned above

You Said:
“The resulting movement of this configuration will be an upward rotation of gyro and lever in vertical plane around the center of hub, i.e. the gyro will arise.”
*If I was going to use or associate the word “precession” anywhere, it would be in describing the motion that you are describing in this sentence. I think this is a language-barriers problem but I am not sure if the problem is yours or mine.

I feel rather strong about using words derived from “precession” in the correct way (as the result and not as the cause), as I have seen it commonly used.
Please let me know what you think about the way the word “precession” is used and what you think its real meaning is.

(I hope my comments are not too harsh, and I will comment on your second posting installment in another reply.)

Best Regards,
Luis

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Answer: Glenn Hawkins - 30/05/2008 02:36:48
 Hi Harry,
Hold Up. You may be right, Harry, and I wrong, but I don’t think so. If you begin extending the axel length, while at the same time adding more main drive force to maintain the same RPMs you would be increasing the speed of the circling gyro, but maintaining the same number in timed tilt rotations in milliseconds that cause lift. The increased speed would cause an increase in momentum certainly, but would that increase centrifuge? I think not. Consider two things, speed and the degree angle of deflection build centrifuge. The greater the radius, the weaker the angle of deflection, the weaker the centrifuge. It looks like lift and centrifuge are locked in a permanent ratio without the possibility of every being manipulated into a variable to provide for a balance point. It seams my earlier assessment was correct, but I’m not sure enough. Give me more time.

Here is a thought problem about a graduated series of G. forces in an odd kind of centrifuge machine for astronauts. Six capsules are attached in outward alinement on the same steel beam. The beam is rotated around a pivot. Do the astronauts, each in his own separate capsule experience the same G. forces, or each a graduated amounts due to varying distances from the pivot? This may not be so easy as it seams. But, if you get it right it may tell you if your theory is probable, or improbable. You will find I know you know, the equation you need in your physic book about the passenger sliding into the outer door as the car goes around a curve.
Take care now Harry,
Glenn


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Answer: Harry K. - 30/05/2008 07:44:35
 Dear Glenn,
Dear Luis,

Thank you very much for your response! Today I'm very busy and therefore I will answer your replies within tomorrow.

Thanks again and best regards,
Harry

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Answer: Harry K. - 31/05/2008 11:48:37
 Dear Glenn,

Thank you for your comments!
To be sure that we talking from the same thing I want to repeat the final claim of my theory, that a lifting force is caused only at the (static) balance point and not during acceleration toward this balance point. When the gyro system reaches the balance point, all acting (vertical) torques are balanced (nullified), but the resultant vector of still acting forces is aligned in an upward direction. Two opposite installed gyros will cause two resultant vectors, which are aligned in upward direction and the resultant vector of these two upward forces will cause a vertical lift.

One comment to your statement regarding extending the axel length:
Please consider, that angular momentum will be calculated by angular velocity and not by RPM (w = 2 * Pi * n). Therefore angular momentum will remain constant if the axel length will be changed, but of course the RPM will change.
Therefore the change of the radius of the rotating gyro around the hub will not change the ratio of tilting force and centrifugal force.
You will see this fact later, when I present the calculations.

Regarding the astronauts in their capsules:
The G-force will increase accordant to the distance of center of rotation (radius), because angular velocity remains constant for all capsules but the centrifugal force will increase with greater radius (Fc = m * r * w * w).

Best regards,
Harry

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Answer: Harry K. - 31/05/2008 12:18:26
 Dear Luis,

Thank you also for your comments! Your comments are important and in no way harsh!

My understanding of "forced precession" is a precession caused by a input torque and not by only gravity. I know this may be misleading, but I thought that this synonym is commonly used here in the correct meaning (refer also to Sandy's explanation: http://www.gyroscopes.org/forum/questions.asp?id=253 ).
However, we can define better synonyms to improve understanding. In future I will talk from "input torque/velocity/rotation" which will cause "forced precession". Agreed?

You Said:
“The precession torque is equal to the torque, caused by forced precession with the drive at the hub.”
*I believe this is the same misnaming issue that I mentioned above

You are right, thank you!
Correction:
"The precession torque is equal to the input torque, caused by the drive at the hub.”

You Said:
“The resulting movement of this configuration will be an upward rotation of gyro and lever in vertical plane around the center of hub, i.e. the gyro will arise.”
*If I was going to use or associate the word “precession” anywhere, it would be in describing the motion that you are describing in this sentence. I think this is a language-barriers problem but I am not sure if the problem is yours or mine.

Again you are right!
Correction:
“The resulting precessional movement of this configuration will be an upward rotation of gyro and lever in vertical plane around the center of hub, i.e. the gyro will arise as a result of precession.”

Thank you, Luis, for your important comments to improve further understanding!

Best regards,
Harry

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Answer: Glenn Hawkins - 31/05/2008 16:25:32
 Dear Harry,

There are some problems.

First let me get this out of the way. Let’s be honest. You said, “I will talk from "input torque/velocity/rotation" which will cause "forced precession". Agreed?” No. That is a terrible idea. Let’s be honest again. Much of one of one above post and all of the other in response can be reduce to: Horizontal force produces vertical precession, or better still simply, ‘upward precession’.

Precession is a 90o torque reaction from applied force of any kind from any direction. Gravity although it has an effect on all things is a completely different matter. Upward precession is an upward torque reaction just as horizontal precession is a horizontal torque reaction. You can use any of the degrees to express this, such as a 45o precession plane, a 30o precession plane etc. and the condition you imply will be quite clearly grasp by your audience. More Later.

Regards, Glenn

Dear Harry,


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Answer: Glenn Hawkins - 01/06/2008 20:03:58
 Dear Harry,

As I understand it, you’re bold and wonderful theory stems from this reasoning: Horizontal rotating forces interact with vertical rotating forces to cause vertical precession, in this case upward lift. Main while centripetal force acts outwardly parallel to the centered main drive motor. Because these two forces, lift and centrifuge are acting at 90o in relation to one another, the assumption is that if they could be adjusted to precisely hold in equal magnitude a resulting force divided between the two would be vectored toward 45o. The theory further assumes that if this position of 45o force could be maintained stably in permanent placement a Balance Point therefore would be established, and from that two opposite gyros so aliened would pull themselves and the entire structure continuously upwards at a 90o angle creating inertial propulsion.

I think the theory, though clever and interesting is flawed as it is based on a Balance Point. I think there can be no Balance Point. I think the mechanics (There’s that blasted word again! right?). I think the mechanics of spinning things will not permit this Balance Point.

This is what I wrote initially:
“If you begin extending the axel length, while at the same time adding more main drive force to maintain the same RPMs you would be increasing the speed of the circling gyro, but maintaining the same number in timed tilt rotations in milliseconds that cause lift. The increased speed would cause an increase in momentum certainly, but would that increase centrifuge? I think not. Consider two things, speed and the degree angle of deflection build centrifuge. The greater the radius, the weaker the angle of deflection, the weaker the centrifuge. It looks like lift and centrifuge are locked in a permanent ratio without the possibility of every being manipulated into a variable to provide for a balance point, but I’m not sure enough. Give me more time.”

Your reply was:
“Therefore the change of the radius of the rotating gyro around the hub will not change the ratio of tilting force and centrifugal force.” You understand perfectly.

Thank you for explaining this, “Regarding the astronauts in their capsules: …angular velocity (RPMs) remains constant for all capsules but the centrifugal force will increase with greater radius…” The question was a little bit tricky and fun wasn’t it, as the capsules were graduated in increasing radiuses distances, while at the same time being forced to rotate at the same speed?

Very good, except your treatment of centrifuge is troublesome to me. I’ll try to explain?

If you rotate a ball bearing towards twenty-five thousand miles an hour the molecule glue will be torn apart by centrifuge long before that speed is reached. Such a bearing under such immense stress in laboratory tests was said to explode, yet you are traveling on the surface of a rotating earth at twenty- five thousand miles per hour and you haven’t exploded all day long, and yet if you collided with a massive steel wall that was in a relatively still position to you, the instant deceleration of the momentum you carry will turn you into a big greasy spot. (Please be careful when you're standing near steel walls, Harry.)

What do you learn from this? Centrifuge is built by increasing revolutions in time as in RPMs, not by increasing momentum.

The earth travels at twenty-five thousand miles per hour. Suppose a ball bearing having a diameter of 1 inch were rotated at twenty-five thousand miles per hour. It would be rotating at 350,139 RPMs. Again, centrifuge would explode the steel bearing, but if you clamped the rotating bearing lightly in a vice it would quickly cease to rotate.

What do you learn from this? Again: Centrifuge is built by increasing revolutions in time as in RPMs, not by increasing momentum.

Having now separated centrifuge from momentum, the theory must assume that centrifuge alone could automatically be increased beyond the power of upward deflection.

So let us now explore my belief in the none-deviational quality of a ratio of force between upward deflection and centrifuge. I predict it can’t be changed into a variate to provide a Balance Point and therefore the theory cannot hold. The question to settle it all is, equations not withstanding, how can this be proven, or disproved?

I am genuinely sorry, Harry if the theory shall be reasoned to fail you.

My Regards, Glenn


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Answer: Harry K. - 02/06/2008 10:44:51
 Dear Glenn,

Thank you for your response. I have not much time but in advance I want to reply to your first paragraph:

I believe you did not get the real function of what I have claimed in my theory.
Horizonatal input rotating forces will cause two effects:

1. 90 deg. precession to the input torque in vertical plane.
2. centrifugal force in horizontal plane.

- These two effects will cause torques around the center of hub in vertical plane.
- The torque caused by precession acts in clockwise (CW) direction, and the torque caused by centrifugal force acts in counter clockwise (CCW) direction.
- The gyro will arise as long as the precession torque is higher than the torque caused by centrifugal.
- If both torques, caused by precession and centrifugal, are equal, then the gyro will remain at this momentary position under a certain angle (not necessarily at 45 deg.!). This is the balance point.
- At this balance point, the torques caused by precession and centrifugal are nullified and therefore the gyro is rotating around the hub in horizontal plane but remains at its position in vertical plane.
- Although the torques are nullified, static forces are still acting! The rsultant force in outward direction and aligned straight with the balance angle will produce sidewise an upward components of this resultant force.
- If 2 gyros are assembled in opposite direction, the sidwise components are nullified and only the upward components are left to cause lift in vertical direction.

I hope you see clearer now. Later more.

Regards,
Harry

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Answer: Glenn Hawkins - 02/06/2008 13:44:07
 Dear Harry,

It is madding that you refuse to understand that I understand and so you just keep repeating yourself. I guess you can’t help it, but if you wished me to take the time and trouble to help you, you must respond to what I said, to all the careful and measured things I said. You cannot rush by ignoring the clarity of my thoughts painfully made simple just for you, saying you are not understood, because you don’t take the time to understood what has been said. Your theory is simple. It is falling down simple. You complicate it with redundancy, unnecessary complications and a strained effort and now you want to add calculations. Why, Harry? One correct little ‘line’ drawing with a few sentences is all that are necessary to explain it. You certainly don’t need calculations to convey this idea. The only complications are in why it does not work, not in conveying the idea of it. So again, I cannot understand why you do not understand that I understand the theorizing. I guess you don’t pay attention. I guess the words sent to you aren’t worth responding to. I guess the thoughts and time freely given to you aren’t even worthy of you asking what they mean if you don’t understand them.

As to your theory the guy, Sandy Kidd and myself have told you in so many words that it doesn’t work and EDH grossed over the possibility of it and went directly to ‘fluttering/overshooting’. Since you’re not responding to what you receive and aren’t taking the time to understand what you read, you I don’t want involvement. You said you did? You want an audience?

None of this is new. Remember the genital suggestion of a languish problem? Remember the humor, but with intent and purpose in, “Hawkins the low grade moron incapable of understanding?” Aw to heck with it. None of us are perfect. You don’t pay attention. You don’t need me. The thing doesn’t work. My effort is wasted. Your mind certainly gets nothing from my effort. From now on it is I who will speed read communications and hurry by unaware of what was meant, while saying I am not understood so I will repeat myself.

By the way I explained your theory for you with clarity, economy and simplification. The only mistake was I said 90o thrust, when I meant 0o. Suggestion: Use my single paragraph explanation if you can discover which it is whenever you wish to be understood, even if you, yourself don’t understand it. You’ll have more success. It is so simply adequate it is much the better.

Regards, Glenn


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Answer: Harry K. - 02/06/2008 14:30:34
 Cool down, Glenn, and drink a beer. This is good for your nerves! Believe me. - But don't go over the top! ;-)

Glenn, I ever stated that this theory is very simple, but although I have the feeling that you do not understand it anyway. This is no criticism to you because I was not able to explain it better. If I write "you do bot understand" it does not mean in any way that I thing you are ignorant!
Oh I hate English language in the meanwhile... :-(

Anyway, nobody forces you to stay here and write comments in this thread of simple and underprivileged theories!

Have a relaxed day!
Harry



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Answer: Luis Gonzalez - 02/06/2008 17:27:30
 Dear Harry,
I read through your 2 main postings and I found the material excellent and I expect it will help greatly in future technical discussions of gyro-propulsion.
I also have found a number of minor errors, and one major error that you may be on the verge of discovering yourself.
I will post when I find a bit of free time.

Best Regards,
Luis

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Answer: Harry K. - 02/06/2008 18:36:54
 Hello Glenn,

Here my detailed comments to what you have "explained your theory for you with clarity, economy and simplification". Thank you for your efforts! :-)))


Glenn wrote:
"Horizontal rotating forces interact with vertical rotating forces to cause vertical precession, in this case upward lift."

No, not at all.
Horizontal rotating forces interact with vertical precession to cause torques which cancel each other. The resulting vector of still acting forces will cause upward lift.
You've got it?

Glenn wrote:
"Main while centripetal force acts outwardly parallel to the centered main drive motor."

No, not at all.
Centripetal force is acting inwards and causes centrifugal force!
You've got it?

Glenn wrote:
"Because these two forces, lift and centrifuge are acting at 90o in relation to one another, the assumption is that if they could be adjusted to precisely hold in equal magnitude a resulting force divided between the two would be vectored toward 45o."

No, not at all.
Cengtrifugal force will cause a torque around the center of hub in vertical plane, which acts in opposite rotation direction as the torque caused by precession around the center of hub in vertical plane.
Depending on all operation data, the gyro will reach and remain at a certain angle. At this position, the torque caused by centrifugal force and the torque caused by vertical precession are EQUAL and therefore BALANCED. The angle MAY be 45 deg. but MAY be any other value between 0 anf 90 degrees, depending from geometric and operation data of the gyro system!
You've got it?

Glenn wrote:
"The theory further assumes that if this position of 45o force could be maintained stably in permanent placement a Balance Point therefore would be established, and from that two opposite gyros so aliened would pull themselves and the entire structure continuously upwards at a 90o angle creating inertial propulsion."

No, but nearly.
The theory further claims, that although the torques are nullified in vertical plane, STATIC forces are still acting. The RESULTANT force will act in an UPWARD direction. This direction depends on the ANGLE of the balance point to horizontal plane of the rotating hub. The greater the angle, the greater the alignment of the resultant force in vertically position.
Two opposite gyros will nullify their force component in sidewise direction and cause therefor 90 degree upward force.
You've got it?

Glenn wrote:
"I think the theory, though clever and interesting is flawed as it is based on a Balance Point. I think there can be no Balance Point. I think the mechanics (There’s that blasted word again! right?). I think the mechanics of spinning things will not permit this Balance Point."

May be you think wrong? It is proved by my comments above, that you did not understand in any way my balance theory, thus your "thoughts" are unimportant in this matter. I would appreciate, if you first try to understand and then provide arguments against my theory, if you find any!

Glenn wrote:
"This is what I wrote initially:
“If you begin extending the axel length, while at the same time adding more main drive force to maintain the same RPMs you would be increasing the speed of the circling gyro, but maintaining the same number in timed tilt rotations in milliseconds that cause lift. The increased speed would cause an increase in momentum certainly, but would that increase centrifuge? I think not."

No, not at all.
I have already tried to explain, that ANGULAR MOMENTUM (L=J*w) will remain constant, if the length of the axle varies! That means, that if you e.g. shorten the axel length, the RPM must be increased to keep the angular momentum constant! Please look in you Physic book, if you own one. Reverse, RPM will be decreased, if the axel length will be extended.
You've got it?

Glenn wrote:
"Consider two things, speed and the degree angle of deflection build centrifuge. The greater the radius, the weaker the angle of deflection, the weaker the centrifuge. It looks like lift and centrifuge are locked in a permanent ratio without the possibility of every being manipulated into a variable to provide for a balance point, but I’m not sure enough. Give me more time.”

It seems you've got it a little bit...

Glenn wrote:
"If you rotate a ball bearing towards twenty-five thousand miles an hour the molecule glue will be torn apart by centrifuge long before that speed is reached. Such a bearing under such immense stress in laboratory tests was said to explode, yet you are traveling on the surface of a rotating earth at twenty- five thousand miles per hour and you haven’t exploded all day long, and yet if you collided with a massive steel wall that was in a relatively still position to you, the instant deceleration of the momentum you carry will turn you into a big greasy spot. (Please be careful when you're standing near steel walls, Harry.)"

Very interesting! I have made a roughly calculation. A 4 inch bearing would rotate with about 2 million revolutions per minute. Wow! No wonder that it will explode. Can you give me more information about the 2.000.000 RPM drive? - Thanks in advance!

Not sure you've got it...

Glenn wrote:
"What do you learn from this? Centrifuge is built by increasing revolutions in time as in RPMs, not by increasing momentum."

Wow, very good! And this circumstance is the reason for establishment of a balance point! I never stated arguments against this fact.

Glenn wrote:
"The earth travels at twenty-five thousand miles per hour. Suppose a ball bearing having a diameter of 1 inch were rotated at twenty-five thousand miles per hour. It would be rotating at 350,139 RPMs. Again, centrifuge would explode the steel bearing, but if you clamped the rotating bearing lightly in a vice it would quickly cease to rotate."

Your calculation is very Interesting! Here is my calculation (sorry, I have to calculate in metric system):
25.000 miles per hour *1.6 *1000 = 40.000.000 meter per hour
Circumference of bearing: 1 inch *25.4 *3.14/1000 = 0.0798 meter
Revolutions of bearing: 40.000.000 / 0.0798 meter / 60 = 8.354.590 RPM (!)

Again, no wonder why the bearing will explode, Glenn! No comment to the quality of your calculations... ;-)

Glenn wrote:
"Having now separated centrifuge from momentum, the theory must assume that centrifuge alone could automatically be increased beyond the power of upward deflection."

No, the theory only claims, that there is a balanced position of the gyro. This position depends on:
- Radius of spinning mass of the gyro,
- Radius of gyro which rotates around the hub,
- angular velocity of spinning gyro,
- angular velocity of rotation around the hub,
- shape factor of gyro (sphere or disk)

Nothing more states this theory.

Glenn wrote:
"So let us now explore my belief in the none-deviational quality of a ratio of force between upward deflection and centrifuge. I predict it can’t be changed into a variate to provide a Balance Point and therefore the theory cannot hold. "

Your belief and predictions are not sufficient. Please write comprehensible ARGUMENTS!

Please read my future postings more carefully before you reply. This would save much time for both of us.

Regards,
Harry

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Answer: Harry K. - 02/06/2008 18:44:19
 Dear Luis,

Thank you very much for your reply. I'm very interested in the errors you may have found.
You know, I also have big doubts about the truth of this theory. I beliefe I'm "blind" to find possible errors by my own and therefor I would be happy if somebody could reveal the errors.

I'm looking forward to your response. However, I'm away for a business trip from Wednesday till Friday.

Best regards,
Harry

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Answer: Glenn Hawkins - 02/06/2008 19:51:28
 Dear Harry,

Listen to yourself. “Anyway, nobody forces you to stay here and write comments in this thread of simple and underprivileged theories!”

Harry, do you think there is somebody here that doesn’t know that? I will not say more as it would be antagonistic. Except if you really must know the truth of it I was actually held here at gunpoint, until just a couple of minutes ago. The assailants went to get coffee.

And. Why was it not clear to you that I had exited this discussion making your later invitation for me to do so a little cockeyed and unnecessary? See if this helps in translation. “You can’t fire me. I quit first.” Oh never mind. It isn’t necessary to understand things like “Elvis has left the building!” Why should you need to understand that means the man has gone and therefore cannot be evicted no mater how much someone might wish to dish it out to him? Therefore…aw forget it. The truth is if you must know, I was forced to stay at gunpoint. Now, you don’t need to understand.

I think you are a nice person. I think you are smart. I think you are probably a good engineer. I like you. Languish? Mine is general clarity, but I will admit it is out of the box from everybody else. Yours is right out of the book—the languish of the engineer. I understand you with difficulty. You can’t, or won’t try to understand me. You still suspect you are not understood, but that at this point is almost incomprehensible to me. Our difficulties are not English and German. It is not even completely standard engineering verse general population clarity. That can be overcome. The problem is that not once in your last two posts have you refereed to anything I said. You see? I would not belittle you intentionally in anyway, in fact I admire you, but you never pay enough attention to what’s said to you. Sometimes you don’t respond to anything at all, but reply anyway. It seems a case of, believing your are not understood and repeating your idea is easier than reading a technical communication multiple times if necessary as the guy has said so many times. Dear Harry he was instructing you though not referring to you by name. I’m sure all I say is wasted, so…

You don’t need me.

Wait a minute. Saying, “You don’t need me.” means I’m exiting this effort, right? I know that’s confusing. Why don’t I just say, “Dear Harry I’ve done all I can in my puny effort to help? Good luck and so long. No wait what I mean is, “Perhaps on other discussions we can meat again, but there’s nothing more I wish to say to this one.” Wait. Wait. I’m sorry, Harry. What I mean to say is I’m not coming back to this post, but maybe another. OK pal? No wait I’m sorry I’m not making myself clear. Why don't you just say for me, “Anyway, nobody forces you to stay here and write comments in this thread of simple and underprivileged theories!”? You see you don’t have to understand me. It’s enough that I understand you. I’m sorry. I will say hello in other posts. Take heed.

Ernest Hemmingway said about another rag, “It is a little like getting booted out of a leper colony.” But, he was exaggerating a little I reckon. So long dear Harry ‘til we meet again.

Sincerely w/regards,
Glenn

I understand, but f#@^# it.









Dear Harry,

Listen to yourself. “Anyway, nobody forces you to stay here and write comments in this thread of simple and underprivileged theories!”

Harry, do you think there is somebody here that doesn’t know that? I will not say more as it would be antagonistic. Except if you really must know the truth of it I was actually held here at gunpoint, until just a couple of minutes ago. The assailants went to get coffee.

I understand! But to heck with it. Believe in contentment however you please. It’s OK. All-is-iss-ok, mox nicks, is-nix-nice-oose! Why do you hate English when spoken in good modern day GI German--drunken American you know? Do you not understand us? Do you have a little sister, yes, yes? Don’t you get mad at me, Harry?

You said, “…underprivileged theories!” Pease believe me I understand this. You feel your theory has been attacked and it has. Such things are personal and can be special to a person and a disagreement can hurt a little. I know. I am sorry. If you noticed my references to your theory in disagreements were handled with greatest care for your feelings. Still and all a disagreement is a disagreement. Be mad at me for that. It’s OK. That might help. I understand. I’m sorry.

I cannot help any more and can only cause more disenchantment. So I will curtsey gracefully and bow out. You have my blessings. “Happy trails to you… until we meet again. Happy trails…” Roy Rodgers & Gabby Hanes sings to Harry K.

Some things are too valuable to lose.
////////////////////////////////////////////////////////////////////////////////////
Harry K. –07/02/3008 17:43:23

Dear Glenn,

"I take this opportunity to apologize. Maybe you will excuse me, please?"

Yes of course, if you accept my apology as well. I think we both were a little bit childish in some matters, therefore we should forget the bad blood. At the end we both have the same target in trying to explain gyro behavior. ;-)
///////////////////////////////////////////////////////////////////////////////////////
:-)

Regards,
Glenn

Now I must stop fooling around and try to make a living though we’re not hurting financially it could happened with a little more bad luck. My wife and I have had cancer, one right after the other. It was a year of indescribable hell before I was diagnosed. I managed to post anyway. We had treatments. She’s recovered a year, I for two months. I have one inch of hair that came back from being blond to black, black. I look funny to me with blue eyes and light completion and jet-black short hair. Joyce came back from straight to curly. Chemo does that and funny things. We were cue ball bald. It fell out in hand fulls. I haven’t been out of the house, but to the hospital in two years except for a little this last month. I’ve had zero exercise. I still look strong, but I’m not. My lungs are very compromised by one of the four serums. This is rare but it happens. Harry, I used to be a supper man. This is true. Only a few knew. At 5’11’ 185 lb. I looked like everybody else who’s not over weight and not overly muscular. I’m not even the man I look like I am. This bothers me all the time. I can feel it. My magic is gone. Time to go to work and stop fooling around. So, so long for a long time Harry. Good luck with your theory. I was never 100 % certain either way.

One last thing, Harry. To your last post, do I got it? I only glanced at some of it. I do not believe you. I think you are just angry. If you honestly believe I am wrong I am bewildered. I am absolutely correct on all counts. How in haven’s name could you be so confused? I can’t believe it. Can I argue? Yes. Should I argue? No. You are just angry. As I said I am leaving. I have no more time for this theory. In the field of engineering try publishing. You will fine yourself alone, one against millions. Your only chance was here with me, away and from your profession and fellow professionals. Good luck to you and again so long. Instead of being frustrated and angry with me, take a few deep breaths and try for more sex. That will make my friend Harry happiest. It’s what your doctor would tell you at this point if he’s any good.


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Answer: Glenn Hawkins - 02/06/2008 20:04:01
 I know it’s time to quit. I get frustrated and make stupid blunders—not in thinking. The first paragraph was not supposed to be sent. It was angry and not friendly and for my eyes only. It is a therapeutic device. ‘Write what you feel then come to your senses and destroy it and begin again more friendlier. I can’t take it back, I’m sorry. You weren’t meant to receive it. I ain’t commin’ back for a long time. It’s my fault.

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Answer: Luis Gonzalez - 02/06/2008 21:27:38
 Dear Harry,

Thank you for laying down a reference framework from which to present our opinions in a somewhat more organized manner. A very lose version of this type of framework has been haphazardly used in many discussions over the years but the lack of discipline has resulted in misunderstanding, and errors that still prevail to this day (despite many attempts to debunk the problems).
Thank you also for your kind response and gracious agreement with my corrections.

I will not go into detail about what you have done correctly because it will require a much longer posting (heaven forbid). Therefore I will go straight into what I see as errors, starting from ordinary errors and then going into the more complex one:

Harry, I don’t mean offense in any way but, please go back and read your last response-posting to me. Note that the way you have written it makes it appear that I am the one saying the statements that I am simply quoting you on; i.e. it makes it appear as if your previous statements are things that I had said, but they were in fact things that I corrected you on because I found them in error.
With all due respect, this forum is a record for posterity and I (gently) object to things that puts erroneous statements in my mouth. I know that being clear is difficult for everyone and I hope you understand why I have to clarify things like this that muddle the record. Please excuse my forthright fullness.

The first and most glaring technical error within your second main posting is in sheet 3 of your drawings; at the bottom there is a set of equations that contains obvious error (in my eyes).
It states: Fr x lr = Fz x lz = Fp x lp - Fz x lz (as a side-note, the use of upper and lower case ‘R’ Vs ‘r’ for the same designator can cause confusion; I interpreted them to mean the same; correct me if I am wrong).
The more glaring error is that there is obviously something missing in this set of equivalences; perhaps you can make your corrections before we evaluate that further.
(I do know that Fz x lz can not be equal to Fp x lp minus the quantity Fz x lz itself, unless Fp x lp is exactly twice as large as Fz x lz).
I also believe there are other problems with these equivalences that should come out in the conversations that follow.

This second main posting of yours introduces an experiment that resembles one of Sandy’s landmark experiments (in Australia I think, using hemispheres) where the hub rotation starts before, and then the spin begins afterwards.
Going straight to the point:
The (“Composite”) Force “Fr”, as derived in your calculations, is in error and does not emerge during experiments (am sure you have already envisioned how to test it). Before I explain the error and how to correct it, I want to say that the force of precessional deflection “Fp” is not a common type of force (am sure we will discuss that eventually).

The absence of the original input torque (the cause) force from your development of equivalences, speaks loudly against the reliability of the conclusions (right from the beginning) because it exclude the basis (cause) of the interactions. It is like trying to build a structure upon thin air.

I will explain (please take a break so you can continue with a fresh mind from here on):

The direction of “Fp” (force “precession”) is in itself the result of combining the INPUT FORCE of the applied torque (“Fi”), and the forces embodied in the motion of “spin” resulting in Fp = Fi(Spin). (Otherwise the causal force “Fi” (of the applied torque) would move the object as it does when there is no spin (deadweight), where there is no 90o deflection.) I hope this is not unnecessarily confusing.

Attempting to combine “Fp” with “Fz” is like mixing apples and oranges; “Fp” IS a 90o response and has a type of complexity that makes it different from “Fz”.
Let me clarify this way:
Mixing “Fp + Fz” is similar to saying that “ay + b = (a + b)y”, which yields incorrect results because “b” dos not have the “y” factor … just as “Fz” does not have the “(Spin)” factor in your equations.
It would be More accurate to say that “ay + by = (a + b)y”…
as well as to say that “Fi(Spin) + Fz(Spin) = (Fi + Fz)(Spin)” because it describes a combination of same type vectors that take into consideration the effects of the spinning object on both forces.
This is just an extension of the rule we agreed about when analyzing “centrifugal force” in the radial configuration, as explained in the thread “EDH's gyro propulsion invention“
At: “http://www.gyroscopes.org/forum/questions.asp?id=887” where we agreed “The centrifugal force causes the spinning-object to turn in the same direction as the hub”.
(By the way this discussion will eventually also lead to the subject we were discussing at the time all focus shifted to your thread on EDH’s invention; at: http://www.gyroscopes.org/forum/questions.asp?id=882.)

**************************
The Most Important thing is to combine the “NET TORQUE” with the “SPIN” factor. This is the most important rule for obtaining correct results about the NET effect(s) of multiple forces interacting with spinning objects etc.
Alternatively you can combine the “SPIN” factor with the force of “ALL INDIVIDUAL TORQUES” beforehand, and then combine the results.
**************************

Though you do not show all your initial steps, it is obvious that your method derives “Fr” as follows:
(1) Fp = Fi(Spin) (90o deflection)
(2) Fr = Fp + Fz
This yields error because “Fz” does not have the (Spin) factor (while “Fp” does).

A correct method is to first combine the forces of all the torques into a net torque force vector, and then use that as the NET applied torque upon the spinning object, as in the following sequence:
(1) Fx = Fi + Fz
(2) Fr = Fx(Spin) (90o deflection)
(“Fx” is just a temporary holding vector that represents the net force of the combined torques.)
The 90o deflection must be applied to the NET resulting vector, after we combine the forces of all the common torques.
You can also apply the 90o deflection to each of the torques individually before combining them and it will yield the same result. However you can not skip applying the deflection to any one of the forces.

In your method you FIRST combine the forces in the (Spin) factor (which causes a 90o deflection), with the input torque (Fi) to produce Fp, and then you treat Fp as a common vector that can be combined with another common force vector in a normal manner… This is an error because Fp contains the (Spin) factor, with the 90o deflection (and Fz does not).
You see, it does matter how you apply the 90o deflection. You can verify this by conducting a couple of mental experiments.
And guess what, my method is congruent with observed results; the spinning-object moves upward (inward…) without stopping at your calculated balance point.
Sandy has been trying to explain this for a long time. It appears to me that Sandy made similar calculation errors in some of his designs a long time ago; that is why his experimental results convinced him that conventional physics does not work with spinning objects. It is an honest error that can happen anyone (including top engineers) while exploring this esoteric area of physics (but it’s still an error).
To make a quick correction to your calculations you must apply the Spin-complex to Fz before combining it with Fp (Fp already has the spin-complex but Fz does not in your calculation).

My explanation is much lengthier than needed (because of how I arrived at my perspective); perhaps you will find a way to state it in the concise way, which you have an excellent talent for Harry. When you do that I would be thankful for the proper and accurate credit, thank you.

Maybe I am deluding my self again but I suspect there are very few people who have given this issue very much thought, except for Sandy. Frankly I am not sure whether anyone else has figured it out (no one has confessed to it, may be afraid to divulge too much).
I suspect Nitro may have an intuitive knowledge of it but his “Nitro’s Law” may have prevented him from delving on it any deeper. That’s one of the problems with having rules of thumb that work well on everything; these rules make it unnecessary to delve much deeper in areas where the rule provides quick answers that are correct.
(I think I’ve said enough for now.)

Best Regards,
Luis

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Answer: Harry K. - 03/06/2008 09:09:16
 Hello Glenn,

No problem, there is no need to appolgize. I know your temperament in the meanwhile. ;-)

Have a great day!
Harry


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Answer: Harry K. - 03/06/2008 09:24:26
 Dear Luis,

Thank you very much for your excellent and productive reply. This kind of communication is that what I'm looking for, thank you!
I have revised the drawings according your notes (Fr instead FR) and of course you are right that the equation on sheet 3 was stated wrong by me (shame on me!). It must be: Fr*lr = Fp*lp - Fz*lz

In the evening I will exchange the documents on my server.
To the other possible errors you have found, I will answer later in a separate post.
In advance, your thoughts are very interesting, but I think there are not correct in this context. I think we have to go much more deeper into detailed explanations, to find out what really happens. But later more.

Best regards,
Harry

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Answer: Harry K. - 03/06/2008 19:20:40
 Dear Luis,

Before I will answer you last posting more detailed, I have prepared a perspective drawing of the gyro setup for better understanding. Here is the link:

http://www.misc.keipert.net/gyro/Gyro_GA.pdf

Is this picture conform with your imagination of this gyro system? Due to your comments I'm not sure if you could have misunderstood some parts of my poor schematic drawings?

Thanks in advance!

Best regards,
Harry

P.S. My business trip was shifted to early next week and thus I can spend more time for the forum. :-)

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Answer: Luis Gonzalez - 03/06/2008 23:19:58
 Thank you for the “perspective” drawing Harry,

Your schematics were sufficient to convey the correct image; as I said, it is a basic configuration that has been discussed many times in this forum. Most gyro propulsion designs use this same basic set-up. There are even explanations and some of the videos also display it though not clearly.

My main point is based on the fact that physics and mathematics go hand in hand.
If a variable is multiplied by a function, the function becomes a factor of that variable, and you can not just add or subtract from it a similar variable that does NOT have that factor.

This is probably the most significant piece of information regarding the physics of spinning objects at this point in time, in this forum. I ask you to evaluate it using your scientific mind (not just engineering). It is common sense.

If you disagree, I would like to know where you see the error and why. Think it through well.
The resulting vector is an upward spiral.

Best Regards,
Luis

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Answer: Luis Gonzalez - 03/06/2008 23:22:33
 Hi Harry,
Correction:
The resulting "Motion" is an upward spiral.
Best Regards,
Luis

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Answer: Glenn Hawkins - 03/06/2008 23:27:34
 Dear Harry,

I see I must come back a last time. About a month ago two people sought to create ways to understand one another better. You were a prime candidate to concentrate on, because your mother tongue is foreign and so a courteous effort was begun to help you learn how to commutate more understandably in this difficult subject. Also your theory was challenged in the process. You became angry. I understand. Your reaction has been, forgive me, beyond reason. You sought ways to attack what is correctly explained. I will go over some of that.

Glenn wrote:
"Horizontal rotating forces interact with vertical rotating forces to cause vertical precession, in this case upward lift."

Harry wrote: No, not at all.
Horizontal rotating forces interact with vertical precession to cause torques which cancel each other. The resulting vector of still acting forces will cause upward lift.
You've got it?

I reply: My statement is simple. In essence I say—“…two forces interact to cause lift.” On this site this is understood to mean as was my statement, vertical gyro spin and horizontal system rotation. I do not say how, where, why, or when, or make an explanation of any kind. I say only-- these two forces interact to cause lift. Now then, you can’t sanely argue contrary to those six words, but you did.

Let us consider: If the two torques cancel one another as you say there is nothing left. End of story. No force remains to do work. What nest? What do you mean ‘the still acting force? Where does this ‘still acting force’ come from? What causes it? What is it?

The meaning of still you'll find here, http://dictionary.reference.com/browse/still There is no meaning in the English languish for what you term as ‘the still acting force’. And ‘canceled’ means to cause to cease to exist. Do you realize you have explained nothing? Why did you challenge me?

I know what you were trying to say. Your, “… the still acting force” you mean is continual. You don’t mean, “Cancel forces” nullified into none existence. You don’t mean that. You mean two forces continually act conjointly to cause a single continuous reaction. You mean ‘act’ ‘act upon’, ‘react’ and simply ‘vector’.

I don’t like doing this to another human who’s English is even less perfect than my own, but then you keep saying to me, “You've got it?” Do I ‘got’ it, Harry? Well yeah. I should hope so. How much simpler could it be? I said, “… two forces interact to cause lift” You tried to correct that? You wrote, “No, not at all.” Were you just angry? Were people being mean to you that day?

Glenn wrote:
"Main while centripetal force acts outwardly parallel to the centered main drive motor."

Harry wrote: “No, not at all.
Centripetal force is acting inwards and causes centrifugal force!
You've got it?”

I reply: I admit this mistake. I offer no defense, no justification. Having said that, you are the only one here, Harry who’s read me who would not understand that I meant Centrifuge, but wrote Centripetal. You are the only one. Do I ‘got’ it? Yeah. I got it, Harry. You have managed to successfully discover a typo error and pounce on it.

Glenn wrote:
"Because these two forces, lift and centrifuge are acting at 90o in relation to one another, the assumption is that if they could be adjusted to precisely hold in equal magnitude a resulting force divided between the two would be vectored toward 45o."

Harry has said: “No, not at all.
Centrifugal force will cause a torque around the center of hub in vertical plane, which acts in opposite rotation direction as the torque caused by precession around the center of hub in vertical plane.

I reply: I haven’t a clue as to what you mean and I have a 145 IQ.

You go on to say: “Depending on all operation data, the gyro will reach and remain at a certain angle.

I reply: “What Data! You mean mathematical calculations? Forget that. Explain how you think it works. I refer you to an expert no longer living, a sweet white haired and kindly man and an excellent teacher, found of saying to his class. ‘If you can’t explain it, you don’t know it.’

Mechanics, Harry. Mechanics explain why and how.

You continue: “At this position, the torque caused by centrifugal force and the torque caused by vertical precession are EQUAL and therefore BALANCED. The angle MAY be 45 deg. but MAY be any other value between 0 and 90 degrees, depending from geometric and operation data of the gyro system! Yeah I got it”

I reply: So you say. So you say. I don’t believe a balance point is possible. That’s the conscientious around here. Forget this data business and explain yourself. Do I ‘got’ it? Yeah I got it, Harry. You’re off the old bean here.

Glenn wrote:
"The theory further assumes that if this position of 45o force could be maintained stably in permanent placement a Balance Point therefore would be established, and from that two opposite gyros so aliened would pull themselves and the entire structure continuously upwards at a 90o angle creating inertial propulsion."

Harry writes: “No, but nearly.
The theory further claims, that although the torques are nullified in vertical plane, STATIC forces are still acting.

Glenn replies: Try to forget the word ‘nullified’. You don’t understand it. You say, “Static forces are still acting?” Do you mean static equilibrium? You suggest I read I read my physics books. I return your lovely intention. http://www.uvi.edu/Physics/SCI3xxWeb/Structure/StaticEq.html
Static forces would not project them selves to cause and action. They remain inert and act only when acted upon. Harry, they would not lift. At best in the words you chose they would hold a position in space whereby a first force could react from them, they being be anchored, but even this is a poor description that can be dismantled.

Harry continues: “The RESULTANT force will act in an UPWARD direction. This direction depends on the ANGLE of the balance point to horizontal plane of the rotating hub. The greater the angle, the greater the alignment of the resultant force in vertically position. Two opposite gyros will nullify their force component in sidewise direction and cause therefore 90 degree upward force.You've got it?”

Yeah, I got it. But I will soon tire of beating on a defenseless person. You shouldn’t have been nasty to me. Even so I’m beginning to feel a little ashamed.

Glenn wrote:
"I think the theory, though clever and interesting is flawed as it is based on a Balance Point. I think there can be no Balance Point. I think the mechanics (There’s that blasted word again! right?). I think the mechanics of spinning things will not permit this Balance Point."

Harry wrote: “May be you think wrong? It is proved by my comments above,”

Glenn replies: My dear Harry nothing is proven. I have argued and so have others for a long time now that mathematical models are not proof, especially when they are based on spinning things. You say, “Components are proven.” This isn’t proof to me. You should know that about me by now. My constant and never ending argument has been for mechanical description and the results of actual testing. These ‘components’ might be argued to some here, but not to others and me here. My face flushes a little when I keep insisting the use of mechanics, that is the explanation of ‘why’ and ‘how’ things happen the way they do, only to see that there is no comprehension and retention of this maddeningly simplest idea in existence. Do as you like. Use your component certainly, but when you address me KNOW ME! KNOW ME! Remember my old teacher’s adherence. “If you can’t explain it you don’t know it.” I think guys understand. I think, thinking ‘why’ and ‘how’ is too hard for them. They were taught from someone else’s mind--mathematics, but not how to think for themselves in terms of ‘why’ and ‘how’. When you try to explain in understandable mechanics you prove your competence level, but in this subject you prove even more clearly what don’t know and are confused about. Have your models about components of spinning things, but it is useless to say to me components are proof. Why don’t you know me by now? I know you! I think explanations about spinning things are too hard for people. That’s why me and another guy begin this month to establish some helpful ways toward explaining. Things like, in this communication you say, “upward precession’, or “vertical precession”. That’s good.

Harry continues: “that you did not understand in any way my balance theory, thus your "thoughts" are unimportant in this matter.

Glenn replies: The theory is child’s play, Harry---until ‘you’ try to explain it. Everyone understands it. This is a lost script for the ‘Three Stooges’? Listen Harry; this isn’t a rubric’s cube of the universe only for geniuses to understand. My dogs running through the woods outside understand this theory. So does the cat. She meows and licked herself to prove she understands your theory. It would be comical that you refuse to believe you are understood when you are told you are understood, except--- you are serious. Maybe it’s simple to me, because I’ve practiced reasoning ‘how’ and ‘why’ for twenty years? Nana. It’s just plain simple. That’s all.

Glenn wrote:
"This is what I wrote initially:
“If you begin extending the axel length, while at the same time adding more main drive force to maintain the same RPMs you would be increasing the speed of the circling gyro, but maintaining the same number in timed tilt rotations in milliseconds that cause lift. The increased speed would cause an increase in momentum certainly, but would that increase centrifuge? I think not."

Harry said: “No, not at all.
I have already tried to explain, that ANGULAR MOMENTUM (L=J*w) will remain constant, if the length of the axle varies! That means, that if you e.g. shorten the axel length, the RPM must be increased to keep the angular momentum constant!

Glenn replies: YOU SEE! THIS IS WHAT GETS ME, HARRY. YOU DON’T PAY ATTENTION TO WHAT’S WRITTEN; YET YOU ARGUE AGAINST IT. THIS IS FROM THE PARAGRAPH DIRECTLY ABOVE. “…while at the same time adding more main drive force to maintain the same RPMs” ---Adding enough drive force will maintain the RPMs. You’ve missed the meaning of that whole paragraph.

I repeat what Harry said above: “I have already tried to explain, that ANGULAR MOMENTUM (L=J*w) will remain constant, if the length of the axle varies! That means, that if you e.g. shorten the axel length, the RPM must be increased to keep the angular momentum constant! Please look in you Physic book, if you own one. Reverse, RPM will be decreased, if the axel length will be extended.
You've got it

Harry this is insulting. In your hurried confusion you don’t understand and you are stubborn and unwilling to understand. You think people here have no mind, or knowledge. This has now reached the point of being crazy. Regulars here have become experts in this field, authorities! You have just instructed me publicly in what, THE SIMPLEST, DUMBEST, STUPIDEST, MOST NEARLY IGNORANT PERSON IN THE CIVILIZED WORLD KNOWS. It is an infuriating accusation. You say TO ME, “Please look in you Physic book, if you own one.” How could you think this simple shiii is not known by everyone? You don’t pay attention. That’s no excuse. How could you think that? Never mind. Are you hitting on all cylinders? Does your elevator go to the top? Are you playing with a full deck?

About a month ago two people sought to create ways to understand one another better. You were a prime candidate to concentrate on, because your mother tongue is foreign and so a courteous effort was begun to help you learn how to commutate more understandably in this most difficult subject to convey. Also your theory was challenged in the process. You became angry. I understand. Your reaction has been beyond reason. You sought ways to attack what is correctly explained.

If anyone should be interested, whatever I've stated as true, is true. Whatever I postulated as probably true is probably true. It is all so simple. I think there's nothing more about this theory I need.

WAIT UP everybody! Good news. I got the wrecked beamer running! It’s missing alternately on three cylinders. I see a reason I think. I can fix that in an hour.

Just as I was finishing this just now, I learn that contracts are coming in without being competed for and without my knowledge. People have learned I might be well. I thought I was finished and the little business shot. I’m happy now. I told you I was a nice person. People like me for that.

Harry, I hope you hate me for a while. Go ahead. Say something nasty. It’s good for your liver. And I’ll feel better---like I deserved it for beating you up. To everybody else, Harry is a knowledgeable man who understands what he is about. Yes he is hurried and impatient and has a different approach than me, but he is struggling in a foreign languish and that accounts for most if not all difficulties. I recommend him highly to you as an especially competent and innovative fellow. We just got in a little fuss, Von Braun and me about rocket science. Kidding, kidding, kidding. Your last drawing I like a lot. Sandy Kidd will be happy too.

I’m out of here. I have my life back. I am one of the fortunate. Bless you all. Bless you all. You too especially, our Spanish el matador. Forgive me.

Glenn


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Answer: Harry K. - 04/06/2008 08:31:38
 Dear Glenn:

"I’m out of here"

Thank you and please do not change your mind!

Thanks,
Harry

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Answer: Harry K. - 04/06/2008 09:23:31
 Dear Luis,

Thank you for the confirmation, I only wanted to be sure.
I know, that this design is a basic set-up of most discussed propulsion designs. Only for clarification, please note, that my theory of a balance point was arised by EDH's design and up to this event I did not believe in achieving propulsion with gyro devices. EDH claims, that he has already achieved lift force with this design, but his explanation how this lift force may occur is very different to my balance theory. Thus my "balance theory" is only an attempt of an alternative explanation why EDH's design may work and how it may produce lift.

I'm also be aware of the fact that variables multiplied by a function nust not be calculated with non multiplied variables.
However I have problems in understanding the context of centrifugal force with spinning mass?
To every point in time centrifugal force is acting at every single spinning mass point. If the gyro is aligned inclined, the outside spinning mass points have an increased centrifugal force and the inside spinning mass points a decreased centrifugal force because the rotation radius of each mass point is different.. However, the sum of all vectors acting at each mass point will be a single resulting vector acting outwards at the center of spinning mass with the same size of centrifugal force as if it would be of non spinning dead weight mass.
This is the same behavior as we have discussed it in your rocket thread. A combination of two different vectors will cause a resultant vector and move the rocket in the direction of the resultant vector.

I think this is the important point and I would be very interested to hear your opinion in advance, why you think that centrifugal force is acting different on spinning mass than it would do on non spinning mass?

Best regards,
Harry


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Answer: Glenn Hawkins - 04/06/2008 12:25:01
 Harry that was not necessary. If you want to be a jerk it can go both ways. I will stay and hound you every time you post if you want to be hatful. Let me return your sentiment. Will you please stop this foolishness, this absurd idea and not return to it? (You like taking it as you gave it.) Sandy Kidd has already built the machine you drew, without the childish presumptions of an impossible balance point. Keep it up with the stupid little remarks meant to be mean. Better for you to shut up when a man says he’s leaving. By the way if you read the critique of your critique that you started you have learned your low competence level. You didn't learned. Unbelievable! You're at it again. Don’t say another word to me.

Today is a great day. Lots of wonderful things to do. I’m happy.


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Answer: Glenn Hawkins - 04/06/2008 14:12:34
 Forget it, Harry. I accept your curt remark. I told you it’d be good for your liver. You have a right to live and be happy. Develop your idea. No laws require you to buy a sense of humor. Tyrrhenum: sapias, vina liques et spatio brevi.

The important things haven’t changes since the first writings of man, 4,700 years ago.

‘Epic Of Gilgamesh’ Siduri: "As for you, Gilgamesh, fill your belly with good things; day and night, night and day, dance and be merry, feast and rejoice. Let your clothes be fresh, bathe yourself in water, cherish the little child that holds your hand, and make your wife happy in your embrace; for this too is the lot of man."
27th century BC

I’m running late, Goodby, Harry and so long. I leave on a good note. Don't spoil it, please.


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Answer: Harry K. - 04/06/2008 19:41:53
 Dear Luis,

We could continue our discussion private via Email, without these parasitic phrases from this poor guy. Please let me know.

Thanks and with best regards,

Harry

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Answer: Harry K. - 04/06/2008 19:45:16
 Dear Mr. Glenn Turner,

would you please so kind and advise Mr. Glenn Hawkins to stop with his unreasonable and impertinent defamations against my person?

If he will not stop with his insulting postings, I ask you to provide his TCP/IP-number.
Please tell him also, that I do not want any further communication with him and all further postings from him will be ignored as from now.

Thanks!
Harry K.

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Answer: Glenn Hawkins - 04/06/2008 22:37:10
 Dear Harry,

You wrote this critique below on me. These are your words. I replied in kind.---------You began this. You started it. Now you can't defend yourself. Now you whine. You don’t want me stop posting in this thread as I said I was quitting. You want somebody to force me.

You don’t get to say the first hateful thing, then say the last hatful thing to a man who said he’s quitting this thread. You began an argument, argued your case, lost your case and now plead for somebody to wipe your bloody nose and do your sniveling biding for you. STAND UP FOR YOURSELF MAN! YOU STARTED IT! and let me part in peace as I’ve asked to ‘please’ to do so. I will prove this at the end of this post.

HARRY WROTE:
Glenn wrote:
"Horizontal rotating forces interact with vertical rotating forces to cause vertical precession, in this case upward lift."

Harry wrote: No, not at all.
Horizontal rotating forces interact with vertical precession to cause torques which cancel each other. The resulting vector of still acting forces will cause upward lift.
You've got it?

I reply: My statement is simple. In essence I say—“…two forces interact to cause lift.” On this site this is understood to mean as was my statement, vertical gyro spin and horizontal system rotation. I do not say how, where, why, or when, or make an explanation of any kind. I say only-- these two forces interact to cause lift. Now then, you can’t sanely argue contrary to those six words, but you did.

Let us consider: If the two torques cancel one another as you say there is nothing left. End of story. No force remains to do work. What nest? What do you mean ‘the still acting force? Where does this ‘still acting force’ come from? What causes it? What is it?

The meaning of still you'll find here, http://dictionary.reference.com/browse/still There is no meaning in the English languish for what you term as ‘the still acting force’. And ‘canceled’ means to cause to cease to exist. Do you realize you have explained nothing? Why did you challenge me?

I know what you were trying to say. Your, “… the still acting force” you mean is continual. You don’t mean, “Cancel forces” nullified into none existence. You don’t mean that. You mean two forces continually act conjointly to cause a single continuous reaction. You mean ‘act’ ‘act upon’, ‘react’ and simply ‘vector’.

I don’t like doing this to another human who’s English is even less perfect than my own, but then you keep saying to me, “You've got it?” Do I ‘got’ it, Harry? Well yeah. I should hope so. How much simpler could it be? I said, “… two forces interact to cause lift” You tried to correct that? You wrote, “No, not at all.” Were you just angry? Were people being mean to you that day?

Glenn wrote:
"Main while centripetal force acts outwardly parallel to the centered main drive motor."

Harry wrote: “No, not at all.
Centripetal force is acting inwards and causes centrifugal force!
You've got it?”

I reply: I admit this mistake. I offer no defense, no justification. Having said that, you are the only one here, Harry who’s read me who would not understand that I meant Centrifuge, but wrote Centripetal. You are the only one. Do I ‘got’ it? Yeah. I got it, Harry. You have managed to successfully discover a typo error and pounce on it.

Glenn wrote:
"Because these two forces, lift and centrifuge are acting at 90o in relation to one another, the assumption is that if they could be adjusted to precisely hold in equal magnitude a resulting force divided between the two would be vectored toward 45o."

Harry has said: “No, not at all.
Centrifugal force will cause a torque around the center of hub in vertical plane, which acts in opposite rotation direction as the torque caused by precession around the center of hub in vertical plane.

I reply: I haven’t a clue as to what you mean and I have a 145 IQ.

You go on to say: “Depending on all operation data, the gyro will reach and remain at a certain angle.

I reply: “What Data! You mean mathematical calculations? Forget that. Explain how you think it works. I refer you to an expert no longer living, a sweet white haired and kindly man and an excellent teacher, found of saying to his class. ‘If you can’t explain it, you don’t know it.’

Mechanics, Harry. Mechanics explain why and how.

You continue: “At this position, the torque caused by centrifugal force and the torque caused by vertical precession are EQUAL and therefore BALANCED. The angle MAY be 45 deg. but MAY be any other value between 0 and 90 degrees, depending from geometric and operation data of the gyro system! Yeah I got it”

I reply: So you say. So you say. I don’t believe a balance point is possible. That’s the conscientious around here. Forget this data business and explain yourself. Do I ‘got’ it? Yeah I got it, Harry. You’re off the old bean here.

Glenn wrote:
"The theory further assumes that if this position of 45o force could be maintained stably in permanent placement a Balance Point therefore would be established, and from that two opposite gyros so aliened would pull themselves and the entire structure continuously upwards at a 90o angle creating inertial propulsion."

Harry writes: “No, but nearly.
The theory further claims, that although the torques are nullified in vertical plane, STATIC forces are still acting.

Glenn replies: Try to forget the word ‘nullified’. You don’t understand it. You say, “Static forces are still acting?” Do you mean static equilibrium? You suggest I read I read my physics books. I return your lovely intention. http://www.uvi.edu/Physics/SCI3xxWeb/Structure/StaticEq.html
Static forces would not project them selves to cause and action. They remain inert and act only when acted upon. Harry, they would not lift. At best in the words you chose they would hold a position in space whereby a first force could react from them, they being be anchored, but even this is a poor description that can be dismantled.

Harry continues: “The RESULTANT force will act in an UPWARD direction. This direction depends on the ANGLE of the balance point to horizontal plane of the rotating hub. The greater the angle, the greater the alignment of the resultant force in vertically position. Two opposite gyros will nullify their force component in sidewise direction and cause therefore 90 degree upward force.You've got it?”

Yeah, I got it. But I will soon tire of beating on a defenseless person. You shouldn’t have been nasty to me. Even so I’m beginning to feel a little ashamed.

Glenn wrote:
"I think the theory, though clever and interesting is flawed as it is based on a Balance Point. I think there can be no Balance Point. I think the mechanics (There’s that blasted word again! right?). I think the mechanics of spinning things will not permit this Balance Point."

Harry wrote: “May be you think wrong? It is proved by my comments above,”

Glenn replies: My dear Harry nothing is proven. I have argued and so have others for a long time now that mathematical models are not proof, especially when they are based on spinning things. You say, “Components are proven.” This isn’t proof to me. You should know that about me by now. My constant and never ending argument has been for mechanical description and the results of actual testing. These ‘components’ might be argued to some here, but not to others and me here. My face flushes a little when I keep insisting the use of mechanics, that is the explanation of ‘why’ and ‘how’ things happen the way they do, only to see that there is no comprehension and retention of this maddeningly simplest idea in existence. Do as you like. Use your component certainly, but when you address me KNOW ME! KNOW ME! Remember my old teacher’s adherence. “If you can’t explain it you don’t know it.” I think guys understand. I think, thinking ‘why’ and ‘how’ is too hard for them. They were taught from someone else’s mind--mathematics, but not how to think for themselves in terms of ‘why’ and ‘how’. When you try to explain in understandable mechanics you prove your competence level, but in this subject you prove even more clearly what don’t know and are confused about. Have your models about components of spinning things, but it is useless to say to me components are proof. Why don’t you know me by now? I know you! I think explanations about spinning things are too hard for people. That’s why me and another guy begin this month to establish some helpful ways toward explaining. Things like, in this communication you say, “upward precession’, or “vertical precession”. That’s good.

Harry continues: “that you did not understand in any way my balance theory, thus your "thoughts" are unimportant in this matter.

Glenn replies: The theory is child’s play, Harry---until ‘you’ try to explain it. Everyone understands it. This is a lost script for the ‘Three Stooges’? Listen Harry; this isn’t a rubric’s cube of the universe only for geniuses to understand. My dogs running through the woods outside understand this theory. So does the cat. She meows and licked herself to prove she understands your theory. It would be comical that you refuse to believe you are understood when you are told you are understood, except--- you are serious. Maybe it’s simple to me, because I’ve practiced reasoning ‘how’ and ‘why’ for twenty years? Nana. It’s just plain simple. That’s all.

Glenn wrote:
"This is what I wrote initially:
“If you begin extending the axel length, while at the same time adding more main drive force to maintain the same RPMs you would be increasing the speed of the circling gyro, but maintaining the same number in timed tilt rotations in milliseconds that cause lift. The increased speed would cause an increase in momentum certainly, but would that increase centrifuge? I think not."

Harry said: “No, not at all.
I have already tried to explain, that ANGULAR MOMENTUM (L=J*w) will remain constant, if the length of the axle varies! That means, that if you e.g. shorten the axel length, the RPM must be increased to keep the angular momentum constant!

Glenn replies: YOU SEE! THIS IS WHAT GETS ME, HARRY. YOU DON’T PAY ATTENTION TO WHAT’S WRITTEN; YET YOU ARGUE AGAINST IT. THIS IS FROM THE PARAGRAPH DIRECTLY ABOVE. “…while at the same time adding more main drive force to maintain the same RPMs” ---Adding enough drive force will maintain the RPMs. You’ve missed the meaning of that whole paragraph.

I repeat what Harry said above: “I have already tried to explain, that ANGULAR MOMENTUM (L=J*w) will remain constant, if the length of the axle varies! That means, that if you e.g. shorten the axel length, the RPM must be increased to keep the angular momentum constant! Please look in you Physic book, if you own one. Reverse, RPM will be decreased, if the axel length will be extended.
You've got it

Harry this is insulting. In your hurried confusion you don’t understand and you are stubborn and unwilling to understand. You think people here have no mind, or knowledge. This has now reached the point of being crazy. Regulars here have become experts in this field, authorities! You have just instructed me publicly in what, THE SIMPLEST, DUMBEST, STUPIDEST, MOST NEARLY IGNORANT PERSON IN THE CIVILIZED WORLD KNOWS. It is an infuriating accusation. You say TO ME, “Please look in you Physic book, if you own one.” How could you think this simple shiii is not known by everyone? You don’t pay attention. That’s no excuse. How could you think that? Never mind. Are you hitting on all cylinders? Does your elevator go to the top? Are you playing with a full deck?

About a month ago two people sought to create ways to understand one another better. You were a prime candidate to concentrate on, because your mother tongue is foreign and so a courteous effort was begun to help you learn how to commutate more understandably in this most difficult subject to convey. Also your theory was challenged in the process. You became angry. I understand. Your reaction has been beyond reason. You sought ways to attack what is correctly explained.

If anyone should be interested, whatever I've stated as true, is true. Whatever I postulated as probably true is probably true. It is all so simple. I think there's nothing more about this theory I need….

….I’m out of here. I have my life back. I am one of the fortunate. Bless you all. Bless you all. You too especially, our Spanish el matador. Forgive me.

Kind regards,
Glenn,

[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[

Harry K. –04/06/2008 08:31:38

Dear Glenn:

"I’m out of here"

Thank you and please do not change your mind!

Thanks,
Harry
[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[

Glenn Hawkins - 04/06/2008 12:25:01

HARRY THAT WASN’T NECESSARY. If you want to be a jerk it can go both ways. I will stay and hound you every time you post if you want to be hatful. Let me return your sentiment. Will you please stop this foolishness, this absurd idea and not return to it? (You like taking it as you gave it?) Sandy Kidd has already built the machine you drew, without the childish presumptions of an impossible balance point. Keep it up with the stupid little remarks meant to be mean. Better for you to shut up when a man says he’s leaving. By the way if you read the critique of your critique that you started you have learned your low competence level. You didn't learn. Unbelievable! YOU ARE AT IT AGAIN. DON’T SAY ANOTHER WORD TO ME.

Today is a great day. Lots of wonderful things to do. I’m happy.

[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[

Glenn Hawkins - 04/06/2008 14:12:34

Forget it, Harry. I accept your curt remark. I told you it’d be good for your liver. You have a right to live and be happy. Develop your idea. No laws require you to buy a sense of humor. Tyrrhenum: sapias, vina liques et spatio brevi.

The important things haven’t changes since the first writings of man, 4,700 years ago.

‘Epic Of Gilgamesh’ Siduri: "As for you, Gilgamesh, fill your belly with good things; day and night, night and day, dance and be merry, feast and rejoice. Let your clothes be fresh, bathe yourself in water, cherish the little child that holds your hand, and make your wife happy in your embrace; for this too is the lot of man."
27th century BC

I’M RUNNING LATE, GOODBYE, HARRY AND SO LONG. I LEAVE ON A GOOD NOTE. DON’T SPOIL IT, PLEASE.

[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[

But you did spoil it—a second time no less.

[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[

Harry K.- 04/06/2008 19:45:16

Dear Mr. Glenn Turner,

would you please so kind and advise Mr. Glenn Hawkins to stop with his unreasonable and impertinent defamations against my person?

If he will not stop with his insulting postings, I ask you to provide his TCP/IP-number.
Please tell him also, that I do not want any further communication with him and all further postings from him will be ignored as from now.

Thanks!
Harry K.

[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[

Silly man. You started a fight of the mind. You can’t fight. Why don’t you drop it?


Report Abuse
Answer: Luis Gonzalez - 05/06/2008 16:21:51
 Hi Harry,

Let’s give this public forum another chance and see if animosity levels drop.
One method that seems to work (to some extent) is, if we don’t have nice things to say to an individual it’s best to say nothing.

If members have technical contributions, they should make them in customary or agreed ways.
No one can coerce responses from others, except by how interesting their ideas about physics and/or gyros are.

Posting solely for the purpose of disruption should be handled by our host Mr. Glenn Turner using the appropriate website tools.

Let’s see if we can play in this arena as mature men, instead of as unruly children in a sandbox.
If we need to, there are other formats in which to discuss technology.

My Best to All,
Luis

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Answer: Harry K. - 05/06/2008 17:57:36
 Hello Luis,

You are right. Disregard might be the best way in this case. I would appreciate if Mr. Turner could clean my thread from this kind of garbage.

So I'm awaiting your technical related reply to my last posting, Luis. Thanks.

Best regards,
Harry

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Answer: Luis Gonzalez - 05/06/2008 20:20:53
 Hi Harry,
I am thinking of responding in a new thread with a reference to this one (when I find the time). What would you prefer?
Regards,
Luis

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Answer: Harry K. - 05/06/2008 20:58:13
 Hi Luis,

I would prefer if all off-topic postings here would be deleted by the admin, but I know as well that this will not happen.
I have started this thread to discuss the possiblity of a balance point of an overhung gyro system. By now I have introduced the theory of this balance point but not yet the necessary calculations. Therefore I think it would not be a good idea to spilt this thread. However, if you prefer to start a new thread, then do it.

Best regards,
Harry

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Answer: Luis Gonzalez - 06/06/2008 21:19:22
 Dear Harry,

If you want to discuss the rocket behavior lets do it in its original thread so we can keep things manageable in this one.
-
To understand the CONTEXT of “ANY force” (including centripetal/centrifugal) in relation to the spinning mass just ask yourself whether (a) the force creates a torque, and (b) whether the resulting torque tries to changes the spin axis (spin plane) Orientation.
--
Don’t let the relative positions of the mass-points, or the magnitude of the force (including centrifugal), confuse the issue (this will not lead to the right answer because it’s not relevant in what we are looking for).
Remember that it is not the centrifugal force (itself) that behaves different on spinning mass; rather it is the resulting toque (around the PIVOT), that behaves different on a spinning mass (as opposed to a non-spinning mass). Why, because the torque tries to change the spin axis (spin plane) ORIENTATION.
(Will it help to compare the torque effect of gravity, to the torque effect of centrifugal around the pivot?)
-
To find the right answer look at the formulas and ask your self if “Fr” is the SAME when calculated your way, or when calculated my way.
I know the results are DIFFERENT.
I know that both can NOT be correct.
The question is… which one is correct… then you can start chasing the reason why!!
-
I would have never figured it out without Sandy’s explanation of his experimental results.
I still would not have been able to figure it out, if I had adhered to Sandy’s interpretation of those experimental results.
-
Here is a set of mostly “YES’ or “NO” questions that lead towards my conclusion regarding the possibility of your balance point:

Definition:
A function can be anything, from a constant, to a complex transformation that involves lengthy formulas (that may include pages of symbols).

Q1.) Does physics abide by mathematical rules?
My answer to Q1: YES.
Up to now that is true without exception, when the correct equation is used.

Q2.) When does a function (or rule) apply to a variable?
My answer to Q2: A common function applies to all variables that are affected by the influence of that function, because these variables are within the scope of the function (e.g. the same “Pivot”), and are not excluded by any specific exceptions.

Q3.) Is the (“Precession-Like”) 90o deflection considered to be a “function” (or rule)? When does it apply?
My answer to Q3: YES.
The “Precession-Like” rule is a consistent formula that, when applied to the direction of a torque in a constant manner, affects any torque that occurs around the PIVOT-POINT 360o X 360o (except for a torque that matches exactly with the axis and direction of the spinning object).

Q4.) Is the applied torque “Fi x li” subject to (affected by) the influence of the “precession-like” rule (function)?
My answer to Q4: YES.
We know that the effects of the applied torque (“Fi x li”) are modified by the “Precession-Like” rule which deflects it by 90o, yielding precessional motion; “f(d)(Fi x li) = (Fr x lr)”. (Where “f(d)” is the “Precession-Like” 90o deflection function.

Q5.) Are torques resulting from centrifuge “Fz x lz” (due to hub rotation) subject to (affected by) the influence of the “Precession-Like” rule (function)?
My answer to Q5: YES.
Centrifuge torque occurs around the PIVOT POINT (and can never coincide with the spin rotation because the pivot point is always aligns with to the spin axis), as the pivot point is attached to that axis. (therefore it tries to modify the spin orientation.)

Q6.) Does the “Precession-Like” function affect BOTH (Fz x lz) and (Fi x li)?
My answer to Q6: YES.
Because of QA4 and QA5 above

Q7.) Are (Fz x lz) and (Fi x li) both subject to mathematical rules?
My answer to Q7: YES.
Because of QA1 above

Q8.) Does “Fr x lr” include (contain) the “precession-like” function?
My answer to Q8: YES.
Because Fr x lr = f(d)(Fi x li); where f(d) is the precession-like deflection function, which causes the 90o response to the torque “Fi x li”.

Q9.) Does “Fz x Lz” include (contain) the “precession-like” function f(d)?
My answer to Q9: NO.
“Fz” is created by the spin but it does not reflect the effect of a 90o deflection until the mass is spun in f(d)(Fx) etc.

Q10.) Can “Fz x lz” and “Fi x li” be combined by addition or subtraction?
My answer to Q10: YES.
They are like terms.

Q11.) Can “Fz x lz” and “Fr x lr” be combined by addition or subtraction?
My answer to Q11: NO.
They are NOT like terms. (However f(d)(Fz x lz) may ve combined with (Fr x lr).)

Let’s get beyond this point and see if we can go on to understand propulsion.
-
-
Here is an interesting question:
Is f(d)(Fi + Fz) equal to (f(d) Fi) + (f(d) Fz)?

If “Yes”, how can we prove it?
The real meaning behind these questions is whether 2 torques, which have been (90o) deflected, can cause each other a new “Precession-Like” deflection when they meet at an angle? Yes or no?

If the answer to this last question is “NO”, then we have found a way to resolve the uncertainty introduced by “Nitro’s Law”.
It also means we may be able to find a way to create lift without falling into the unwanted motions from the spinning object (as it responds to other forces and obstacles).
I believe this is what EDH’s device has succeeded in accomplishing.

Best Regards,
Luis

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Answer: Harry K. - 07/06/2008 15:49:35
 Dear Luis,

Again, thank you very much for your productive reply, which highlights your very good mind!
I'm pretty sure you are aware, that you have confirmed most of my claims stated in context with my balance theory? I think, we both try to explain our beliefs from different point of views, and that is very good. If we both can convince each other from our point of view, we are a big step ahead!

You wrote:
"To understand the CONTEXT of “ANY force” (including centripetal/centrifugal) in relation to the spinning mass just ask yourself whether (a) the force creates a torque, and (b) whether the resulting torque tries to changes the spin axis (spin plane) Orientation."

Thank you for clarification. I thought you have believed that the CAUSE of centripetal/centrifugal forces could be different for spinning and non-spinning masses.
Your notes under a) and b) describe the principle of the claims in my balance theory:
- Centrifugal force causes a torque, which interacts with precessional deflections of a spinning gyro. In this context, the EFFECT of centrifugal force has an influence to the spin axis of a gyro in precession. That's exactly what I claim in my balance theory but Sandy claims that this is impossible.

You wrote:
"Remember that it is not the centrifugal force (itself) that behaves different on spinning mass; rather it is the resulting toque (around the PIVOT), that behaves different on a spinning mass (as opposed to a non-spinning mass). Why, because the torque tries to change the spin axis (spin plane) ORIENTATION.
(Will it help to compare the torque effect of gravity, to the torque effect of centrifugal around the pivot?) "

Yes I agree and I don't know when I have stated different claims?
I think, at this point we have to go deeper in gyro behavior and repeat some facts.

First we have to point out the differences between regular precession and forced precession.

REGULAR PRECESSION
1) Imagine a non-spinning overhung gyro, aligned and fixed in horizontal plane. If the non-spinning gyro would be released, it would move downwards around the pivot in vertical plane in exactly the same way as normal dead weight mass. Gravitation causes a torque in vertical plane (Fi x li). This caused torque will accelerate the mass (gyro) until the mass reaches either the ground or the bottom dead center. During acceleration of the mass (gyro), the torque caused by gravitation decreases, because the lever arm decreases continual till it is zero and thus the torque will be zero as well at the bottom dead center. Vertical movement will stop here.

2) Now imagine the same initial situation but now with a spinning gyro. If the spinning gyro would be released, the same torque as before caused by gravitation and lever arm tries to accelerate the gyro to move it downwards in vertical plane. But because the gyro mass is spinning now, this torque will be deflected about 90 degree and accelerates the gyro in horizontal plane to a certain constant velocity around the pivot. Important is the fact, that the initial torque (Fi x li) does not cause ANY movement in vertical plane but in horizontal plane. This initial torque is STATIC but cause MOVEMENT in 90 degree deflected plane. This movement in horizontal plane would never stop, as long as gravitation would act on the gyro, and if friction forces would not be considered.

Please excuse the repeat of these well-known facts, but I think it's important for the next steps.

FORCED PRECESSION
3) Now imagine the same situation as described in 3), i.e. the overhung gyro precesses around the pivot in horizontal plane, caused by the torque Fi x li created by gravity.
Now the caused precession movement in horizontal plane will be constrained by something, for example by friction in the pivot. This friction causes a resistance in horizontal movement around the pivot. This resistance will cause a counter torque (Ff x lf) to precessional torque (Fi x li) in vertical plane.
This counter torque will be again deflected back about 90 degrees, but now in vertical plane. The resulting torque in vertical plane can be calculated by (Fi x li) - (Ff x lf). That means, that the calculated torque, based on the smaller velocity around the pivot by friction, should be smaller than the real acting torque caused by precession. But because this is NOT the case, the spinning gyro will move downwards in vertical plane. This gyro system is now unbalanced and therefore the gyro moves downwards until it reaches either the ground or the bottom dead center. The greater the resistance torque in horizontal plane, the faster the downward movement in vertical plane.

4) Forced precession is a similar behavior of a gyro as explained in 3). A spinning gyro will be prevented by something to precess regular. This prevention in regular precession may be caused by a mechanical construction (something similar as in Sandy's devices) or by any other applied torques for instance counter torques caused gravitation or CENTRIFUGAL forces. Thus, if there are acting ANY counter torques against regular precession, these counter torques will be deflected again about 90 degrees and cause a movement in the same plane as the initial acting input torque (Fi x li). That means that the beforehand STATIC acting input torque (Fi x li) will be accelerated by counter torques, which are acting against precession to "90 deg. deflected precession" to precession. Therefore the input torque is "forced" to move by "deflected precession" instead to remain static, as it would be the case in regular precession.

These statements may be sound complicated and difficult to understand, but the reason is my poor explanation in English language. I hope you and others understand it anyway.

To summarize, a moving, non-static acting input torque is caused by "backward" acting precession, created by torques, acting against precession. These counter torques to regular precession may be caused by e.g. gravitation, centrifugal, or any other applied forces or toques.

And now, if you remember, we have theoretical reached again the issue "egg or hen":

- At regular precession (horizontal STATIC input torque), there are no centrifugal forces present and therefore a counter torque to regular precession cannot be established.
Thus the input torque (Fi x li) must remain static (no movement in horizontal plane) and will be deflected about 90 deg. in vertical plane and causes the gyro to move upwards around the pivot.
Therefore a balance point cannot exist in this situation, because FORCED precession cannot occur by centrifugal forces.
(By the way, please note that words in capital letter are used to underline statements and not in any way negative mind).

To come along with the "egg and hen" problem, I have suggested to change the initial operation conditions to achieve a counter torque caused by centrifugal forces:
1. First start the rotation around the pivot in horizontal plane with non-spinning gyro.
2. Afterwards the gyro begins to spin and increases its rotation slowly to its final velocity.

Under these operation conditions, a balance point should be established, because centrifugal forces are present all the time.
However, I'm really not sure if this balance point may help to achieve propulsion. But I'm really sure, that this balance point is true.

I'm looking forward to your comments. I hope we are now much closer in our thoughts.

Best regards,
Harry





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Answer: Harry K. - 07/06/2008 16:16:34
 DEAR LUIS,

PLEASE IGNORE MY PREVIOUS POSTING. I HAVE CORRECTED SOME MISTAKES IN STATEMENTS AND TYPOS.

THANKS
HARTY


Dear Luis,

Again, thank you very much for your productive reply, which highlights your very good mind!
I'm pretty sure you are aware, that you have confirmed most of my claims stated in context with my balance theory? I think, we both try to explain our beliefs from different point of views, and that is very good. If we both can convince each other from our point of view, we are a big step ahead!

You wrote:
"To understand the CONTEXT of “ANY force” (including centripetal/centrifugal) in relation to the spinning mass just ask yourself whether (a) the force creates a torque, and (b) whether the resulting torque tries to changes the spin axis (spin plane) Orientation."

Thank you for clarification. I thought you have believed that the CAUSE of centripetal/centrifugal forces could be different for spinning and non-spinning masses.
Your notes under a) and b) describe the principle of the claims in my balance theory:
- Centrifugal force causes a torque, which interacts with precessional deflections of a spinning gyro. In this context, the EFFECT of centrifugal force has an influence to the spin axis of a gyro in precession. That's exactly what I claim in my balance theory but Sandy claims that this is impossible.

You wrote:
"Remember that it is not the centrifugal force (itself) that behaves different on spinning mass; rather it is the resulting toque (around the PIVOT), that behaves different on a spinning mass (as opposed to a non-spinning mass). Why, because the torque tries to change the spin axis (spin plane) ORIENTATION.
(Will it help to compare the torque effect of gravity, to the torque effect of centrifugal around the pivot?) "

Yes I agree and I don't know when I have stated different claims?
I think, at this point we have to go deeper into gyro behavior and repeat some facts.

First we have to point out the differences between regular precession and forced precession.

REGULAR PRECESSION
1) Imagine a non-spinning overhung gyro, aligned and fixed in horizontal plane. If the non-spinning gyro would be released, it would move downwards around the pivot in vertical plane in exactly the same way as normal dead weight mass would do. Gravitation causes a torque in vertical plane (Fi x li). This caused torque will accelerate the mass (gyro) until the mass reaches either the ground or the bottom dead center. During acceleration of the mass (gyro), the torque caused by gravitation decreases, because the lever arm decreases continual until it is zero and thus the torque will be zero as well at the bottom dead center. Vertical movement will stop here.

2) Now imagine the same initial situation but now with a spinning gyro. If the spinning gyro would be released, the same torque as before, caused by gravitation and lever arm, tries to accelerate the gyro to move it downwards in vertical plane. But because the gyro mass is spinning now, this torque will be deflected about 90 degree and accelerates the gyro in horizontal plane to a certain constant velocity around the pivot. Important is the fact, that the initial torque (Fi x li) does not cause ANY movement in vertical plane but in horizontal plane. This initial torque is STATIC but cause MOVEMENT in 90 degree deflected plane. This movement in horizontal plane would never stop, as long as gravitation would act on the gyro, and if friction forces would not be considered.

Please excuse the repeat of these well-known facts, but I think it's important for the next steps.

FORCED PRECESSION
3) Now imagine the same situation as described in 2), i.e. the overhung gyro precesses around the pivot in horizontal plane, caused by the torque Fi x li, created by gravity.
Now the caused precession movement in horizontal plane will be constrained by something, for example by friction in the pivot. This friction causes a resistance in horizontal movement around the pivot. This resistance will cause a counter torque (Ff x lf) to precessional torque (Fi x li) in horizontal plane.
This counter torque will be again deflected back about 90 degrees, but now in vertical plane. The resulting torque in vertical plane can be calculated by (Fi x li) - (Ff x lf). That means, that the calculated torque, based on the now smaller velocity around the pivot (caused by friction), should be smaller than the real acting torque caused by gravitation. But because this is NOT the case, the spinning gyro will move downwards in vertical plane. This gyro system is now unbalanced and therefore the gyro moves downwards until it reaches either the ground or the bottom dead center. The greater the resistance torque in horizontal plane (friction in this case), the faster the downward movement in vertical plane.

4) Forced precession is a similar behavior of a gyro, as explained in 3). A spinning gyro will be prevented by something to precess regular. This prevention in regular precession may be caused by a mechanical construction (something similar as in Sandy's devices) or by any other applied torques, for instance counter torques caused by gravitation or CENTRIFUGAL forces. Thus, if there are acting ANY counter torques against regular precession, these counter torques will be deflected back again about 90 degrees and cause a movement in the same plane as the initial acting input torque (Fi x li). That means that the beforehand STATIC acting input torque (Fi x li) will be accelerated by counter torques, which are acting against 90 deg. deflected precession. Therefore the input torque is "forced" to move by "deflected precession" instead to remain static, as it would be the case at regular precession.

These statements may be sound complicated and difficult to understand, but the reason is my poor explanation in English language. I hope you and others understand it anyway.

To summarize, a moving, non-static acting input torque is caused by "backward" acting precession, created by torques, acting against regular precession. These counter torques to regular precession may be caused by e.g. gravitation, centrifugal, or any other applied forces or toques.

And now, if you remember, we have theoretical reached again the issue "egg or hen":

- At regular precession (horizontal STATIC input torque), there are no centrifugal forces present and therefore a counter torque to regular precession cannot be established.
Thus the input torque (Fi x li) must remain static (no movement in horizontal plane) and will be deflected about 90 deg. in vertical plane and causes the gyro to move upwards around the pivot.
Therefore a balance point cannot exist in this situation, because FORCED precession cannot occur by centrifugal forces.
(By the way, please note that words in capital letter are used to underline statements and not in any way negative mind).

To come along with the "egg and hen" problem, I have suggested to change the initial operation conditions to achieve a counter torque caused by centrifugal forces:
1. First start the rotation around the pivot in horizontal plane with non-spinning gyro.
2. Afterwards the gyro begins to spin and increases its rotation slowly to its final velocity.

Under these operation conditions, a balance point should be established, because centrifugal forces are present all the time.
However, I'm really not sure if this balance point may help to achieve propulsion. But I'm really sure, that this balance point is true.

I'm looking forward to your comments. I hope we are now much closer in our thoughts.

Best regards,
Harry


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Answer: Luis Gonzalez - 09/06/2008 00:42:39
 Posted
Dear Harry,

Thank you for your kind words, and please also Note that I use capital letters only to underline.

We are still grappling with the basic issues but I sense we are making some progress (if we were mind readers this process would progress much quicker…oh well).
I am asking you to please respond to my Q1 through Q11, so we can be certain where we stand in relationship to those items. Thanks.

Keep in mind that (gentle) “Disagreement” is just as important as “Agreement”, in the path to mutual recognition of a set of truths (but not if either is done for their own sake).
True meeting of the minds occurs without compromising what we truly think.

Regarding your # 1), I agree with all said but will try to use (Fg x lg) for gravity torque, instead of (Fi xli) to differentiate, and avoid confusion to other readers (no big deal).

Regarding your # 2), I agree, and again would use (Fg x lg) instead of (Fi xli).

Regarding your #3), I have some issues with the language used, and one main item of disagreement.
A) First the language issue:
Harry said:
“This resistance will cause a counter torque (Ff x lf) to precessional torque (Fi x li) in horizontal plane.”
** To call (Fi x li) a precessional torque creates an ambiguity, which later comes back to confuse the true interaction of the forces involved.
“Fi x li”, which is better called (Fg x lg) in this case, can not be horizontal, so it must be (Ff x lf) that is horizontal…that is what your sentence says…
(Note that (Fi x li) is simply the input torque…, and the term “precessional torque” should be reserved for another concept that will emerge in a future conversations.)
B) Now, the item I disagree with:
The combination of “gravity-torque” and “resistance/friction-torque” yields a new resulting torque that may have a different magnitude, but most IMPORTANT has a different DIRECTION ( (Fg x lg) + (Ff x lf) = (Fx x lx) ).
After we apply the 90o deflection function f(d) to the combined torque (Fx x lx), the resulting direction is in a downward DIAGONAL direction; it traces a DOWNWARD SPIRAL (it is a horizontal-downward motion).
** You see, Focus on the new TORQUE DIRECTION of (“Fx x lx”) yields the correct result, without need to introduce the concept of UNBALANCE; the result of f(d)(Fx x lx) is a balance in a new direction (the resulting motion).
** The intuitive expectation that a SMALLER velocity of precession will result, is not well founded on fact because it intuitively attempts to subtract the horizontal-plane “Resistance” force (“Ff”), from the horizontal-plane Precession (“Fp”), (because they occur in opposite directions).
However, the correct combination/interaction of (“Ff x lf”) with (“Fi x li”) is NOT in OPPOSITE directions; it is closer to orthogonal. Therefore a combination of the two torques (Fi x li) + (Ff x lf) = (Fx x lx) is a more accurate representation (better than subtracting one from the other because “Ff” and “Fi” (or “Fg”) are not opposite forces). Please review my Q1 trough Q11 to see how they fit in.

Regarding your #4), I have some issues with the way the term “counter torque” is used but I will not address that here to prevent confusion in this thread.
I do agree that “Forced Precession” is similar in behavior to your item#3 (though there are differences to be discussed elsewhere), and I agree in principle with the basic comparison; however my objections from item #3 carry over to item #4.
Also, I CANNOT agree that (Fi x li) will be accelerated by “counter torques” because these 90o deflections are NOT (Fi x li) input torques (by definition); they only occur in the same direction as (Fi x li) (I will reserve further discussion on this until we have resolve the other differences, and have found out your perception about my Q1 through Q11; so we can both be in “the same page”).

Regarding the “egg and hen” problem (as you call it), the experiment you describe in steps 1, & 2 (at the end of your posting), have been discussed in this forum since Sandy presented this exact experiment, with VALUABLE results, about 2 years ago in this forum (this is a must-read to gain better understanding, though you must NOT get tangled on his INTERPRETATION of the results).
I agree that normal “centrifugal” force (WITHOUT TORQUE!) exist before gyro-spin is introduced.
But, very soon after the gyro-spin is introduced, the “Centrifugal Force” becomes a vertical (inward) “STATIC centrifugal TORQUE”, in a similar manner that you described “static input torque” earlier regarding regular precession. Since the centrifuge is STATIC, it CANNOT be used to participate in the type of balance that you are referring to (because its effective motion is in another direction, at 90o).

The Q1 through Q11 points provide a solid theoretical basis, and the results from Sandy’s experiments provide initial proof that under “Mechanically Forced Precession”, the force from the centrifuge becomes “static” (in the sense that you have introduced “static”) in the same way that the force of gravity becomes static on a gyroscope configuration.
(The proof is also in experiments that can be conducted with limited monetary expenditure.)

Our rate of progress is not very fast but I think it provides “difficult-to-see” answers in response to many questions that are based on human intuitions, and those intuitions are tainted by normal perception of interactions among non-spinning objects.
There is value to what we are doing, but there is greater value in the actions of those who are building models pioneering the transportation methods of the future, whether they fully understand or not.
The sooner we unravel this technology, the sooner it will become common knowledge to all who dare use it.

I will wait for your responses to the 11 questions and for your thoughts on this response.

Best Regards,
Luis

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Answer: Harry K. - 09/06/2008 20:56:11
 Hello Luis,

I'm currently in Egypt for business. Please be patient and wait for my answers till end of this week. Thanks.

Best regards,
Harry

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Answer: Luis Gonzalez - 09/06/2008 22:12:23
 Have fun in Egypt Harry.
Best Regards,
Luis

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Answer: Luis Gonzalez - 13/06/2008 16:34:37
 Dear Harry,

Point of clarification:
I very much like the concept of “STATIC FORCES” that you introduced to describe what happens in cases where resulting motion is deflected (though the point may be argued).
I believe that the words “static-force” can become useful in many ways, especially when describing interactions and behaviors involving spinning objects.

To start, we may say that static-forces (vectors) can be added amongst themselves but CANNOT be added to normal forces (vectors).
This is just another way to state the important and interesting RULE that I have been championing in my recent postings.

I know the rule makes sense but I am not sure whether anyone else is grasping it, and understand its significance to maintaining integrity within the laws of motion.
Without the rule people have been expecting tremendous amounts of “inertial propulsion” during interactions, where none exists.
Applying the rule correctly removes the error in calculation and we can continue working on propulsion without thinking that the laws of motion have been compromised; correct rules and standards serve to improve calculations and to demystify unusual phenomena.

Best Regards,
Luis

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Answer: Harry K. - 14/06/2008 12:05:06
 Dear Luis,

Thank you for your reply. First I will comment your questions Q1-11 with this posting. Later I will answer your posting from 09/06/2008 00:42:39.
One comment in advance regarding your comment of the "rate of progress". I think we are discussing here the very basics of gyro behavior and this necessary, because most of the readers here are not aware of this basics. Therefore the "rate of progress" is sufficient enough or may be too fast, because I'm afraid we still think different in some very important issues regarding basic gyro behavior. So you should not complain the "rate of progress" in every of your postings until we both are convinced and agree with the results of the way of thinking of each other.

Regarding you questions:
Q1.) Does physics abide by mathematical rules?
My answer to Q1: YES.
Up to now that is true without exception, when the correct equation is used.

*** Yes of course! I'm a technical designer and this is my staff of life!

Q2.) When does a function (or rule) apply to a variable?
My answer to Q2: A common function applies to all variables that are affected by the influence of that function, because these variables are within the scope of the function (e.g. the same “Pivot”), and are not excluded by any specific exceptions.

*** A "variable" is certainly a part of the function. A "variable" can take every possible value of a beforehand defined range of values and therfore you are not right with your comment "...and are not excluded by any specific exceptions".

Q3.) Is the (“Precession-Like”) 90o deflection considered to be a “function” (or rule)? When does it apply?
My answer to Q3: YES.
The “Precession-Like” rule is a consistent formula that, when applied to the direction of a torque in a constant manner, affects any torque that occurs around the PIVOT-POINT 360o X 360o (except for a torque that matches exactly with the axis and direction of the spinning object).

***Why do you use "Precession-Like"? - There is precession or there is no precession but nothing between.
I agree that 90 deg. deflection is considered to be a function. I disagree with "(except for a torque that matches exactly with the axis and direction of the spinning object).",
The torque must not match exactly with the axis of the spinning object. The axis of the spinning object can be located at every PARALLEL distance to the axis of pivot!

Q4.) Is the applied torque “Fi x li” subject to (affected by) the influence of the “precession-like” rule (function)?
My answer to Q4: YES.
We know that the effects of the applied torque (“Fi x li”) are modified by the “Precession-Like” rule which deflects it by 90o, yielding precessional motion; “f(d)(Fi x li) = (Fr x lr)”. (Where “f(d)” is the “Precession-Like” 90o deflection function.

*** I disagree. The "input torque" Fi x li will not be affected by 90 deg. deflection!
Input torque Fi x li = precession torque Fp x lp ! The deflection function F(d) causes only the size of precession velocity of the deflected input torque Fi x li = Fp x lp.
This understanding is very important for any further following steps of gyro behavior and therefore please think again in detail about this issue!

Q5.) Are torques resulting from centrifuge “Fz x lz” (due to hub rotation) subject to (affected by) the influence of the “Precession-Like” rule (function)?
My answer to Q5: YES.
Centrifuge torque occurs around the PIVOT POINT (and can never coincide with the spin rotation because the pivot point is always aligns with to the spin axis), as the pivot point is attached to that axis. (therefore it tries to modify the spin orientation.)

*** In general I agree but I don't agree with your statement: "...because the pivot point is always aligns with to the spin axis), as the pivot point is attached to that axis."
The pivot point is NOT always aligned with the spin axis of the gyro, because the hub rotation is always in HORIZONTAL plane but this is NOT the case for the arising gyro! Therfore centrifugal torques vary between maximum (gyro in horizontal alignment) and zero (gyro in vertical alignment).

Q6.) Does the “Precession-Like” function affect BOTH (Fz x lz) and (Fi x li)?
My answer to Q6: YES.
Because of QA4 and QA5 above

*** I disagree. Centrifugal torque Fz x lz is not affected by the input torque Fi x li but only by the hub rotation velocity and the momentary position of the precessing gyro.

Q7.) Are (Fz x lz) and (Fi x li) both subject to mathematical rules?
My answer to Q7: YES.
Because of QA1 above

*** Yes of course! What a question! ;-)

Q8.) Does “Fr x lr” include (contain) the “precession-like” function?
My answer to Q8: YES.
Because Fr x lr = f(d)(Fi x li); where f(d) is the precession-like deflection function, which causes the 90o response to the torque “Fi x li”.

*** I disagree. Please refer to my answer of Q4. The resultant torque Fr x lr is affected only by the input torque Fi x li and centrifugal torque Fz x lz because input tprque = output (precessional) torque: Fi x li = Fp x lp#

Q9.) Does “Fz x Lz” include (contain) the “precession-like” function f(d)?
My answer to Q9: NO.
“Fz” is created by the spin but it does not reflect the effect of a 90o deflection until the mass is spun in f(d)(Fx) etc.

*** It's not really clear to me how you understand this question. The centrifugal torque is insofar affected by precession function f(d) because f(d) affects the hub rotation velocity and thus the size of centrifugal force.
Please explain your understanding to this question more detailed.

Q10.) Can “Fz x lz” and “Fi x li” be combined by addition or subtraction?
My answer to Q10: YES.
They are like terms.

*** Yes, because FI x li = Fp x lp, i.e. Fz x lz and Fp x lp can be combined by addition or subtraction.

Q11.) Can “Fz x lz” and “Fr x lr” be combined by addition or subtraction?
My answer to Q11: NO.
They are NOT like terms. (However f(d)(Fz x lz) may ve combined with (Fr x lr).)

*** I don't agree. Fr x lr stands for the vectorial addition of Fz x lz and Fp x lp Therefore the approach of your question is wrong.

So far to the response of your questions Q1 - Q11. I will answer your other postings later, may be by tomorrow. I would appreciate if you could wait with your response till I have answered your other postings. Thanks in advance!

Best regards,
Harry


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Answer: Harry K. - 15/06/2008 14:17:59
 Dear Luis,

Here is my answer to your posting from 09/06/2008 00:42:39.

Regarding your comment to #1) and #2) with your advice to use Fg x lg instead of Fi x li.
I think the use of Fi x li is more general but I have no problems with the use of Fg x lg instead of Fi x li.

Regarding your comment to #3)
Re A) You are right of course. It must be read as:
"This resistance will cause a counter torque (Ff x lf) to precessional torque (Fp x lp) in horizontal plane." Precession torque Fp x lp as well as the friction torque Ff x lf are acting in horizontal plane in this context.

Re B) Please note that the friction torque works against precession torque in this context. Friction torque against initial input torque Fi x li or Fg x lg is not considered in this context. Sorry if I have caused confusion in this matter.
Thus in this context there is a lower precession torque (Fp x lp) - (Ff x lf) and this causes a 90 deg. "back deflection" to the inital tilting torque Fg x lg with the effect, that the axis of the gyro will precess with lower velocity and tilt as well in vertical plane. The tilting velocity (back deflection) is according the function F(d) based on the size of friction torque Ff x lf. If the axis of the gyro will move in horizontal plane (precession) AND vertical plane (back deflection), then the gyro system is unbalanced, because the gyro is constrained in its precession and is forced to counter act against this disablement.
May be you see in every point of time the momentary resulting direction of the gyro, but I see always two different 90 deg. movements in horizontal and vertical plane, because one movement in a plane is always the reaction of an initial movement in the other (90 deg.) plane and reverse. I hope you understand what I mean.

Regarding your comment to #4)
The agreement that counter torques are at the end the reason for forced precession is mandatory for further discussions. If not any counter torques to precessiion torques would act in a gyro system, no input angular velocity would occur. The input torque Fi x li would remain STATIC, i.e. STATIONARY without any movement!
Also, BECAUSE these counter torques are acting in the same direction of the input torque (because they were deflected back from precession plane), the input torque Fi x li BEGINS to move!

Please, I must insist on your agreement regarding this basic issue, otherwise further discussion will not make sense for both of us.

Regarding "egg and hen" issue:
It doesn't matter whether issues were discussed here already in the board or not. If I have understood Sandy correctly in the EDH-thread, the gyros in his device were fixed or adjustable by mechanic constructiions and therefore his valuable results may not be comparable to the EXPECTED results of the design which we are discussing here.

Please note, that the introduced balance point is the result of only STATIC forces. The frame of reference is rotated synchronous with the rotation around the hub. Please note, that a moving frame of reference at constant velocity is same STATIC as a non-moving frame of reference! That means that the forced movement of the input torque is not relevant for any static calculations. However, the forced movement is necessary for the cause and size of centrifugal force, but not for afterwards static calculations!
This fact is also very important to understand my theory of the balance point.

Our rate of progress is indeed not very fast. There are many differences in basic understanding yet, but I hope and believe that we will overcome these differences sometime. How fast? - I don't know... ;-)

Best regards,
Harry



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Answer: Luis Gonzalez - 15/06/2008 14:22:11
 Dear Harry,

If we can’t agree about the basics, how can we agree about anything else?
Can we agree about Truth or Falsity on 2 basic rules (to find the simplest common ground)?

Basic Rule #1:
The INPUT Torque “(Fi x li)” occurs in a HORIZONTAL Plane (as a motor applies this torque upon the hub).
I say TRUE, how about you?

Basic Rule #2:
Any PRECESSIONAL Torque “(Fp x lp)” produced, would occurs in a VERTICAL Plane (because the precessional motion moves up & inward).
I say TRUE, how about you?

If both these basic rules are True, then “(Fi x li)” and “(Fp x lp)” are vectors in different directions.
And vectors that point in different directions CANNOT be equal, therefore (Fi x li) cannot equal (Fp x lp).

You say: “Input torque Fi x li = precession torque Fp x lp !”
I say they are vectors pointing in different directions and therefore cannot be equal.
Can we ever get past this point?

-
The tone of your posting tells me you are unhappy about something.
I find the sharp remarks discourteous, and I do not accept that behavior from a conversation partner.
I can’t do very much when insults comes from individuals I don’t talk to, but in a dialog we must be address each other in a civil manner. When I have nothing nice to say, I prefer to say nothing.

Thank you,
Luis

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Answer: Harry K. - 15/06/2008 14:33:56
 Dear Luis,

Here my answer to your posting from 13/06/2008 16:34:37:
Certainly one should be able to calculate vectors correctly. A force is always a vector and therfore vectors have to be calculated always geometrically by considering of value AND direction of vectors.
A standard procedure for correct calculations of vectors is to calculate the individual X- and Y- components of each vector seperately. Torques, however, can be calculated easily by standard addition or subtraction rules.

These are standard mathematic rules and I wonder if you really think that somebody here could not be aware of these rules?

Best regards,
Harry


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Answer: Harry K. - 15/06/2008 14:55:09
 Dear Luis,

Sorry, it is not my intention to be unfriendly in any way. However, it must be possible in our conversataion to point out issues clearly without phrases.

To your rules:#

"Basic Rule #1:
The INPUT Torque “(Fi x li)” occurs in a HORIZONTAL Plane (as a motor applies this torque upon the hub).
I say TRUE, how about you? "

*** I also say TRUE.

"Basic Rule #2:
Any PRECESSIONAL Torque “(Fp x lp)” produced, would occurs in a VERTICAL Plane (because the precessional motion moves up & inward).
I say TRUE, how about you?"

*** I also say TRUE.

"If both these basic rules are True, then “(Fi x li)” and “(Fp x lp)” are vectors in different directions.
And vectors that point in different directions CANNOT be equal, therefore (Fi x li) cannot equal (Fp x lp).

You say: “Input torque Fi x li = precession torque Fp x lp !”
I say they are vectors pointing in different directions and therefore cannot be equal.
Can we ever get past this point? "

***No, Fi x li and Fp x lp are TORQUES but no VECTORS! Only the plane of rotation of these TORQUES are different.
If (Fi x li) would not be equal to (Fp x lp), you could deflect a part of the precession torque back to the input and design a independent loop without any input torque!. Or you could drive a generator with a part of the higher precession torque and build a Perpetuun Mobile. ;-)

I hope this helps a little bit and please appologise if you think my tone was unfriendly.
By the way, I'm very happy! :-)

Have a nice Sunday!
Harry


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Answer: Luis Gonzalez - 15/06/2008 17:43:18
 Dear Harry,

Thank you for the nice wish, and I hope you enjoy your Sunday too.
At least we agree about the 2 basic rules, but we may still be a long way from a meeting of the minds.

We have arrived at an arguable point; whether torques should be considered to be an angular type of vector or not. (I have seen torque represented as a vector in some physics books and can see the point.)
This should make for an interesting discussion in another thread, but maybe not here.

However, we do know that a Force VECTOR is one of the 2 factors represented by a Torque.
I say this is true, how about you?

In my opinion the equal sign (“=”) is reserved for physical and mathematical equality.
You appear to be using the equal sign (“=”) where only equality of magnitude may exists (which is also not necessarily the case).

In the subject of spin interactions, which we are discussing, magnitude plays a limited role, because the orientation of a force can reduce or increase the net effect dramatically.
We can discuss this at length if you wish because this may be one of the main reasons for disagreeing.
I CANNOT agree that (Fi x li) = (Fp x lp) because the force factor in each one of these equations is a VECTOR in a different direction and can produce different effects in their interaction with (other) 3rd forces.

Further, even the magnitude of input torque (Fi x li) does NOT need to be equal to the magnitude of precession torque (Fp x lp).
More accurately the magnitude of the input torque (Fi x li) only needs to be GREATER than, or equal to the magnitude of the precession torque (Fp x lp), in order to prevent the “Perpetual Motion” problem.
In other words, if (Fp x lp) is SMALLER than (Fi x li) then there is no energy coming out from nothing; the problem only occurs if we were to allow (Fp x lp) to be larger (Fi x li), which cannot occur.

You see, looking only at the wrong half of the picture results in erroneous conclusions, such as saying that the input torque “Fi x li” must be equal to the precession torque “Fp x lp”, which is not necessarily true.
I hope this helps in seeing the true picture.

Best Regards,
Luis

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Answer: Harry K. - 15/06/2008 21:06:13
 Dear Luis,

Thank you for your quck reply. You are right, torques can be represented as torque vectors, but this is only a definition for differentiation. I have already stated that input torque and precession torques are acting in different planes. If you want to call the different planes "vectors" then it is okay for me. When I state Fi x li = Fp x lp then I mean the values are equal but of course the planes of these acting torques are different.

For consideration of the possibilty of a balance point, only the values of the torques and the rotational direction in the accordant plane are important because the calculation is based on static forces in a "frozen" status, i.e. without movement.

Of course it is true that a force VECTOR is one of the 2 factors represented by a torque. However the force vector is changes always its DIRECTION.

If you don't agree that the value (or magnitude) of the torques Fi x li and Fp x lp are equal, then please present your calculation of the magnitude of precession torque. To state that the precession torque may be smaller than the input torque without any calculation or explanatory statement, is not enough.

I'm not sure who of us can really see parts of the true picture? Maybe none of us!? You think regarding the balance point in motions but I think in static equilibrium without motion. My way of thinking is the way how I had learnt it and how I calculate stress, torques and forces acting on machine parts.
I don't know yet how to come closer in our individual points of view? But I still will not give up. Maybe it would be a good idea to stop here for a while and to continue first with my calculations of the (possible) balance point? What do you think about this idea?

Best regards,
Harry

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Answer: Luis Gonzalez - 16/06/2008 14:31:16
 Dear Harry,

I will respond to your item “4)” from your posting of 07/06/2008 15:49:35, as soon as I find some time, because I also think it is important.

Best regards,
Luis

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Answer: Luis Gonzalez - 16/06/2008 14:36:12
 Hi Harry,

It may be a good idea to take a break and work on your claculations.
I will repond to your item "4)" becaus that is still hanging heavily.

Best Regards,
Luis

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Answer: Luis Gonzalez - 21/06/2008 16:28:01
 Dear Harry,

First let me clarify what I mean by “f(d)”:
The function “f(d)” is simply short way of saying that a 90o transformation has occurred TO a force vector. (F(d) = Transform a force vector by 90o.)

-
Notes about input (tilting) torque and resulting motions:

#-I) When the input torque (Fi x li) is “STATIC” it creates NO centrifuge (zero centrifuge), which means that (Fi) is fully converted into precession-motion / force.
In other words the PROPORTION of input-force (Fi) that is deflected is near or equal to 100%, under a “static” input torque.

#-II) On the other hand, when the input torque is able to produce any hub-system rotation (i.e. “Fi x li” is NOT “STATIC”), then some PROPORTION of the input hub-force becomes converted into the hub-rotation, which then produces CENTRIFUGE.
*This means that ONLY the remaining PROPORTION of input force (Fi) can become converted into (upward/inward) deflection (i.e. NOT 100% of “Fi x li” becomes precession “Fp x lp”).
(Important Note: A “restraint” must be introduced (temporarily?) to allow the initial occurrence of hub-rotation, while PREVENTING precession from concluding the trial run before centrifuge can be created)

Conclusion about input (tilting) torque and resulting motions:
-We know that the input (tilting) torque can be made to drive (a) the upward/inward (primary) deflection, and also (b) the horizontal hub-rotation; and we also know that force is required to drive either of these 2 motions.
-We may envision different ways (or paths) in which the hub-rotation comes to exist, but it is ultimately derived from the force of the input (tilting) torque (the mathematics can derive the same results through these different means of calculation).

-Bottom Line:
*The upward/inward (primary) deflection, and the horizontal hub-rotation are DIFFERENT motions; they occur in different directions and with different magnitudes.
*Since both motions are derived from the same input (tilting) torque “Fi xli”, then this is a known situation where the input torque does NOT equal to the (primary) deflection torque.
**I think this should serve to provide the proof you requested Harry. It proves that (Fi x li) is NOT necessarily EQUAL to (Fp x lp) under certain known conditions.

--
Regarding your item “4)”:
After re-reading your item “4)”, I have found a significant problem with the overall line of reasoning.
a.) In item “4)” you entertain the effect of forces that prevent (or interfere with) regular precession’s motion / force.
b.) You highlight CENTRIFUGAL force as one such interfering force, and correctly reiterate the fact that an interfering force or centrifuge will deflect back around into the same plane (and direction) as the acting (input) torque.
c.) Then you say that the “STATIC” input torque (Fi x li) will become accelerated by the deflected CENTRIFUGE… *** Perhaps you have spotted the gleaming error by now…
*** It’s clear that a “STATIC” input torque CANNOT produce a CENTRIFUGE…, as stated in my item “#-I” and “#-2” above. “Static” input torque and the type of centrifuge that we are talking about, are mutually excluded.

The bottom line is that, if the input torque is expected to produce CENTRIFUGE, then it must first be able to produce HUB-ROTATION (i.e. “Fi x li” must NOT be “STATIC”). Further, as I stated in an earlier posting, under normal conditions the velocity of “deflected-centrifuge” is always slower than the velocity of the hub-rotation (a rotation which is necessary to create the centrifuge).

Does the slower velocity of the deflected-centrifuge require modification to the mathematics used in describing simultaneous interactions among centrifuge, hub-rotation, and object-spin?

I hope you find this information helpful.

---
To continue in a hopefully productive analysis:
The existence of CENTRIFUGAL force necessitates the existence of a CENTRIPETAL force, and this is also holds true under our (much discussed) hub model where the input torque is NOT “STATIC” but rather ACTIVE (because the hub turns).
Based on this assumption, here are some questions (some with obvious answers) that may help to bring light to some issues.

Q1a. Where does the centrifuge force come from?
A2a. The force of the centrifuge is derived from the hub-rotation that occurs during “ACTIVE” input hub-torque (i.e. “non-static” hub-torque), when the pivot arm forms an angle greater than 0o.

Q2a. Where does centripetal force come from?
A1a. Above 0o, centripetal force is supplied by the force of “precession”, which results from the input hub-torque.
(Note that precession-force, produced with ‘static’ input torque, never becomes a centripetal tether because there is no centrifuge in that type of interaction).

Q3a. What proportion (or ratio) of the input hub-torque converts to “precession” (which then supports centripetal force)? Vs.
Q4a. What proportion (or ratio) of the input hub-torque supports conversion into centrifugal force? (Note that even though the CENTRIFUGE force itself is “STATIC”, the hub torque needs to be active; i.e. “Fi” should NOT be “STATIC”.)
A3a & 4a. The proportion of input hub-torque that converts to precession/centripetal force Vs centrifuge depends on the RATIO of “spin” to “hub-rotation”, though it is not a linear equation (perhaps a very clever person can derive the correct equation).
Also, other physical constraints and forces can affect this ratio.

----
Sandy conducted many of experiments.
The particular experiment that I referred to is one where Sandy started the hub-rotation first, and then afterwards, began the spin from very slow, and then increasing spin gradually.
He did NOT include any additional physical/mechanical constrains on this experiment, and his results have been of great value to this forum.
I will provide a link to his explanation of the experiment when I find the correct forum location.

The unexpected results of these experiments, along with some unorthodox views about what occurs in a gyroscope, lead Sandy to conclude that there is “no mass” after the “Saturation” zone occurs.
I think you have managed to explain that it is the force (not the mass) that is transferred down through the axis of the system rotation.

----
I would like to start a discussion about why I think deadweight mass is important to obtain a balance, and present some math & physics that I believe proves it.

Best Regards,
Luis

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Answer: Harry K. - 22/06/2008 15:01:42
 Hello Luis,

Thank you for your feedback. It gave me the opportunity to learn more about your way of thinking.
Luis, one point in advance: any by me stated comments or explanations regarding precession behavior, sizes of input and output torques, etc. are based on my own experiments and logical thinking. I made these experiments in the late eighties and they were related more to energy transfer issues than to gyro propulsion issues. So you can be sure that I have thought very deeply regarding the basics of gyro behavior.
I also could present integral functions to calculate the precession velocity of each mass point during 0 and 180 degree to an acting tilting torque. I have this all done already in the late eighties with the help of my former ATARI computer. This integral function proves the equation for precession wP = T / LG (wP... angular velocity of precession, T... tilting torque, LG... angular momentum of spinning gyro mass).

Please excuse my digression but I want that you be aware that I had dealt with this stuff intensive from the past till now.

Regarding f(d):
Yes I know the sense of this function! ;-)

You wrote:
"#-I) When the input torque (Fi x li) is “STATIC” it creates NO centrifuge (zero centrifuge), which means that (Fi) is fully converted into precession-motion / force.
In other words the PROPORTION of input-force (Fi) that is deflected is near or equal to 100%, under a “static” input torque. "

*** Yes, I'm totally in agreement with your statement.

You wrote:
"#-II) On the other hand, when the input torque is able to produce any hub-system rotation (i.e. “Fi x li” is NOT “STATIC”), then some PROPORTION of the input hub-force becomes converted into the hub-rotation, which then produces CENTRIFUGE.
*This means that ONLY the remaining PROPORTION of input force (Fi) can become converted into (upward/inward) deflection (i.e. NOT 100% of “Fi x li” becomes precession “Fp x lp”). "

***Basically I agree with your statement, however, your way of thinking is not consistent. WHY is the input torque be able to produce any hub-system rotation? - You do not present any explanation of this very important behavior!
My answer is: The input torque is able to produce hub rotation only then, when the precession of the gyro will be constrained by any counter force (better: counter torque) against precession!
For example, if a counter torque to precession would be 50 percent of the size of the input torque, the result would be that precession velocity would decrease to 50 percent as well, AND the input torque would begin to rotate with (the remaining) 50 percent as well! In this case, precession velocity and hub rotation velocity would have exactly the same size.
There is another important point. If the hub rotation begins, angular momentum in the hub rotation plane is the result. The necessary energy for accelerating the gyro in rotation plane is stored in angular momentum and will be transferred back to the gyro system if the input torque decreases or will be stopped. The same energy in form of angular momentum is stored in the precession plane as well. If the gyro shape behave equal in both planes (input and precession plane), e.g. a spherical shape, energy stored in angular momentum is identical for both rotation planes (input and precession plane).
I have discussed this issue already in this thread: "http://www.gyroscopes.org/forum/questions.asp?id=864"

Also you should be aware, that during "full precession", i.e. the input torque is static and precession velocity is at 100 percent, there occur no physical work at the input. But as soon as a counter torque to precession causes the rotation of the input torque at the hub, physical work will occur in the plane of the input torque. So if you constrain or block the gyro in precession plane by mechanically design, you must induce physical work into the input plane: P = Ti x wi (P... mechanical power, Ti... input torque, wi... angular velocity of input torque).
You see, the input torque will not change, only the angular velocity will change according the size of the counter torque which constrains the precession of the gyro.
If the gyro will be blocked in its precession (this is the case at the balance point), the precession velocity will be deflected totally back to the input plane and the result is, that the gyro rotates with unchanged input torque and precession velocity, but there is not any movement left in precession plane because this movement was blocked.

Now, if you accelerate the gyro system with blocked gyros in precession plane, e.g. by a motor drive, the input torque as well as the hub velocity will be accordingly increased. That also means, that the now static torque, which blocks the gyro in precession plane, will be increased in the same way. So it can be understood again, that the torque, which blocks the gyro in precession plane, causes the input torque of the motor drive.

Again, you are right that there will occur a distribution of rotation velocities between input and precession plane, but this has no influence of the size of the input torque and deflected torque, the sizes of this torques are always equal. The understanding of this fact is essential for general understanding of gyro behavior.

You wrote:
"(Important Note: A “restraint” must be introduced (temporarily?) to allow the initial occurrence of hub-rotation, while PREVENTING precession from concluding the trial run before centrifuge can be created) "

***As explained before, a CONTINUING "restraint" must be applied to allow the occurrence of hub-rotation! To achieve the influence of centrifugal, the hub must rotate BEFORE the gyro mass is spinning!

You wrote:
"Conclusion about input (tilting) torque and resulting motions:
-We know that the input (tilting) torque can be made to drive (a) the upward/inward (primary) deflection, and also (b) the horizontal hub-rotation; and we also know that force is required to drive either of these 2 motions.
-We may envision different ways (or paths) in which the hub-rotation comes to exist, but it is ultimately derived from the force of the input (tilting) torque (the mathematics can derive the same results through these different means of calculation)."

***You are right, that the input torque is at the end the reason for all following movements. However, the reason for horizontal hub-rotation is ALWAYS a restraint of precession of a gyro. If there would not be any restraint in precession of a gyro, the input torque would remain static and thus no physical work would be performed in horizontal plane.

You wrote:
"-Bottom Line:
*The upward/inward (primary) deflection, and the horizontal hub-rotation are DIFFERENT motions; they occur in different directions and with different magnitudes.
*Since both motions are derived from the same input (tilting) torque “Fi xli”, then this is a known situation where the input torque does NOT equal to the (primary) deflection torque.
**I think this should serve to provide the proof you requested Harry. It proves that (Fi x li) is NOT necessarily EQUAL to (Fp x lp) under certain known conditions."

*** No, in reality each mass particle moves in unique directions. Because of the design of a gyro, each mass point has another diametric mass point, which moves in counter direction. That's the reason of the 90 degree deflection, but the cause is the gyro design which distributes ONE movement of each mass particles in two orthogonal planes.
As explained before, if a gyro will be blocked to precess, then Fi x li is equal to Fp x lp,
Please note, that this is the case if there is a STATIC balance point. This static balance point is basement for my calculations.


Regarding your comments of my item "4)" you are certainly right. Sorry for this mistake!


You wrote:
"The bottom line is that, if the input torque is expected to produce CENTRIFUGE, then it must first be able to produce HUB-ROTATION (i.e. “Fi x li” must NOT be “STATIC”). Further, as I stated in an earlier posting, under normal conditions the velocity of “deflected-centrifuge” is always slower than the velocity of the hub-rotation (a rotation which is necessary to create the centrifuge). "

*** If a STATIC balance point does exist, and this is what I claim, then CENTRIFUGAL torque works AGAINST precession torque. And the cause of this fact is the EXISTENCE of the hub rotation. I know this claim is an antagonism, and therefore I had mentioned the "chicken and egg" problem in prior postings.
We should discuss this problem later, after we are in agreement with the basics.

You wrote:
"Q1a. Where does the centrifuge force come from?
A2a. The force of the centrifuge is derived from the hub-rotation that occurs during “ACTIVE” input hub-torque (i.e. “non-static” hub-torque), when the pivot arm forms an angle greater than 0o.

*** My answer A1a: The force of the centrifuge is derived from the hub-rotation that occurs by RESTRAINING forces against precession. Centrifugal forces are acting when the pivot arm forms an angle EQUAL OR greater than 0 degree but smaller than 90 degree.
(Note: centrifugal TORQUE exists when the pivot arm forms an angle greater than 0 degree and smaller than 90 degree).

You wrote:
"Q2a. Where does centripetal force come from?
A1a. Above 0o, centripetal force is supplied by the force of “precession”, which results from the input hub-torque.
(Note that precession-force, produced with ‘static’ input torque, never becomes a centripetal tether because there is no centrifuge in that type of interaction)."

*** My answer A2A: Centripetal force is supplied by the RESULTANT force, caused by centrifugal and precession.
Please note, that the pivot arm is only able to transfer tensile or compressive forces. The resultant force acts therefore always in the momentary direction of the pivot arm.

You wrote:
"Q3a. What proportion (or ratio) of the input hub-torque converts to “precession” (which then supports centripetal force)? Vs.
Q4a. What proportion (or ratio) of the input hub-torque supports conversion into centrifugal force? (Note that even though the CENTRIFUGE force itself is “STATIC”, the hub torque needs to be active; i.e. “Fi” should NOT be “STATIC”.)
A3a & 4a. The proportion of input hub-torque that converts to precession/centripetal force Vs centrifuge depends on the RATIO of “spin” to “hub-rotation”, though it is not a linear equation (perhaps a very clever person can derive the correct equation).
Also, other physical constraints and forces can affect this ratio. "

*** My answer to A3a % 4a: I have already explained the coherences regarding precession, input torque and centrifugal. I believe I'm not a very clever person, but I'm sure to be able to derive the correct equations. This is certainly an important part of my calculations. ;-)


I would be happy if you could find the link of his explanations!

You wrote:
"I would like to start a discussion about why I think deadweight mass is important to obtain a balance, and present some math & physics that I believe proves it. "

Deadweight mass supports to achieve a balance point, but it have negative influence of propulsion because of the additional mass in the system. A balance point could also be achieved, if a gyro would pivot around its center of spinning mass and if a deadweight mass would be fixed at the end of one rotation axle.
Before you start a new thread I would wait for presenting my calculations.

However, I'm very surprised about your change in mind, that you now believe in a balance point and not in a pulsating device described by EDH? ;-)

Best regards,
Harry



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Answer: Luis Gonzalez - 24/06/2008 02:13:43
 Hi Harry,

I read over your response quickly and it looks very good; your arguments appear clear and well presented.
I will need more time to read it slower for a well thought out response.

Best Regards,
Luis

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Answer: Luis Gonzalez - 03/07/2008 17:52:12
 Hi Harry,
Thank you for the great posting.

Fist of all, I am sure we would not be having these conversations if we did not both know that we have spent years thinking and investigating this subject. Unfortunately, it is still easy to underestimate others and that goes both ways.
On the good side, I see that we are now in agreement in a good number of items and I want to list the ones that I am currently aware of.

Before I do that, allow me to provide meaning to the symbols I intend to use:
(Fi) = Input force
(Fg) = Force of gravity
(Fp) = Force of precession
(Ff) = Force that interferes or resists against precession
(Fz) = Centrifugal force
(Fh) = Force of hub-rotation

ITEMS WE AGREE UPON:

Agreement #1: - The magnitude of (Fi x li) never changes (except when the input motor’s power is increased), even if angular velocities change for other reasons (such as obstacles etc).

Agreement #2: - (Fi x li) is the sole cause for all system movements in the hub model (except for the spin motion).

Agreement #3: - Without an interfering obstacle or force (Ff), the input torque (Fi x li) or (Fg x lg) is “Static” and converts only into (Fp x lp) (friction not included).

Agreement #4: - Some form of (Ff x lf) must exist, to enable hub-rotation (which can then result in centrifuge)

Agreement #5: - When (Ff x lf) interferes with (Fp x lp) then only a portion of (or none of) (Fi x li) continues to convert into (Fp xlp), and the remaining portion of (Fi x li) converts into hub-rotation

Agreement #6: - Centrifugal force (Fz) results from the hub-rotation and causes centrifugal torque (Fz x lz) when the pivot arms are between 0o and 90o

Agreement #7: - Centrifugal torque (Fz x lz) deflects into the direction of EXISTING hub-spin (when the sphere or flywheel is spinning).

Agreement #8: - Centrifugal force (Fz) can only occur when a restrain or force (Ff) acts against precession (and enables hub-rotation).

Agreement #9: - In absence of material tensile-radial strength (between 0o and 90o), the inward/upward force of precession (Fp x lp) provides the centripetal force (-Fz), but the precession torque (Fp x lp) is greater than the centrifugal torque (Fz x lz), for the specific ratios of spin to hub-rotation that we are addressing. (Whenever the centrifugal torque (Fz x lz) is equal to the precession torque (Fp x lp), then a balance point can exist. In my opinion, this balance point can only occur with some additional help, as I have stated, and will elucidate on one of my next postings.)

Agreement #10: - If the interfering torque (Ff x lf) becomes equal to the precession torque (Fp x lp), then upward/inward precession stops; this is necessary for a “balance point” to occur.

Agreement #11: - The angular momentum stored in the motion of precession is very small in proportion to other potentials in the system.

Agreement #12: - Though we have somewhat different ways of determining the effects of multiple forces and torques within the hub-environment, our methods are congruent and arrive at the same basic results (we will see how it evolves going forward).

Agreement #13: - I believe we are in agreement that every torque introduced into the hub-system must have its net-effects transposed by 90o (“f(d) function”).

These items are what I see as our points of agreement up to now, and I hope this list continues to grow as we work towards a meeting of the minds.

Next time I have an opportunity to spend time on my hobby, I will post a number of items that we need to resolve but don’t yet have agreed about.

Best Regards and I will be back soon with a completion to my response,
Luis

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Answer: Harry K. - 08/07/2008 20:13:05
 Dear Luis,

Sorry for the late reply but´I was busy. Thank you for the great job of your list of agreement! I have noticed that you have changed your mind in some important issues and therefore I'm happy that we are now in agreement with many basic gyro behaviors!

Please allow me some additional notes:

Re agreement #1:
The magnitude of (Fi x li) can change if the gyro arises from its horizontal plane, because the gyro leaves the (always) horizontal plane of the applied input torque (Fi x li). That means, the input torque has its maximum at 0 deg. of the gyro position, and has its minimum (zero) at 90 deg. of the gyro position.
However the relation between input torque and precession torque will not change, i.e. important for calculations is the position of the gyro, where centrifugal and precessional torque are balanced (balance point). At this point the input torque can be accurate calculated.

Re agreement #5:
Your definition is in my opinion too spongy. It is important to understand that the hub-rotation will be only caused by a (deflected) counter torque (Ff x lf) to precession torque (Fp x lp). Therefore the correct definition should be:

"When (Ff x lf) interferes with (Fp x lp) then only the difference torque of (Fi x li) - (Ff x lf) continues to convert into (Fp x lp). The deflected torque (Ff x lf) will be converted into hub-rotation.
You see, if (Ff x lf) = 0 then precession rotation = maximum and hub-rotation = 0.
If (Ff x lf) = (Fp x lp) then precession rotation = 0 and hub-rotation = maximum.
Between these both extremes, precession and hub rotation are devided accordingly and the input torque will always be equal to (Ff x lf) or (Fp x lp) if the gyro is blocked in precession direction. In this case both torques have the same size: (Ff x lf) = (Fp x lp) and therefore the input torque has the same size as well.
Once you have full understand this important behavior, all following understanding in gyro behavior will be much easier. ;-)

Re agreement #8:
"In absence of material tensile-radial strength (between 0o and 90o), the inward/upward force of precession (Fp x lp) provides the centripetal force (-Fz),"

I do not agree with this statement. The centripetal force is provided by the design of the lever. The lever is free to move up- and downwards but forced to rotate around the center of hub by the hub drive. The forced circular path of the lever provides the centripetal force but not the force of precession!

I agree with the rest of your statements. In my opinion the balance point can simply occur, if all gyro parameters are well defined and if the gyro will be started to spin SLOWLY AFTER hub-rotation has established by the hub drive.

Re agreement #11:
This depends on the gyro parameters. If the the precession velocity will be high, the stored angular momentum may be high as well. However you are right that under normal conditions the precession velocity will be much slower than the spinning velocity of the gyro itself.


Thank you again for your fine arranging of our basic agreements. And I'm also happy that we are able to communicate with the defined and accepted variables and terms! IT's now much easier to understand each other!

Best regards,
Harry

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Answer: Luis Gonzalez - 08/07/2008 21:16:46
 Hi Harry,
As I work on my response to your previous-to-last posting, I find myself sorting through specifics that I have been avoiding for a ling time.
Please excuse my delay, as there are items I need to think through before completing my response.

Best Regards,
Luis

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Answer: Luis Gonzalez - 14/07/2008 22:26:13
 Dear Harry,

Congratulations on very productive postings.
I started to write a point to counterpoint response but decided to avoid answers that may not clarify issues.

I want to introduce (Fq x lq) = f(d)(Fz x lz) as a secondary precession derived from centrifuge.

I also want to expand on “Fh”, the force of the hub-rotation, where:
(Fh x lh) + (Fp x lp) = (Fi x li) + (Ff x lf).
This equivalence equation helped me to understand what you really meant when, in your own way, you said that (Fi x li) has the same magnitude as (Fp x lp)…
To become convinced, first I noticed that when (Ff x lf) completely blocks precession, then (Ff x lf) must have become equal in magnitude to (Fp x lp), and both are locked into NON-ACTION… At this point (Fh x lh) also becomes equal to (Fi x li), and one may say they are in a unique form of balance, because the 4 torques now appear to have the same magnitude.
This equivalence equation started me thinking in a productive direction thanks to you Harry.

I believe that I am now able to see why (Fi x li) has the same magnitude as (Fp x lp), and therefore why (Ff x lf) must have the same magnitude as (Fh x lh); though I admit it is not something I could easily figure out on my own, and I still have difficulty seeing and accepting (not exactly intuitive).
My confusion comes from a couple items. One confusing factor is the role played by the rate of spin. The other source of confusion is from thinking about “Velocities” instead of “Forces” and “Torques”.
Thank you for getting me to think-through this very important point.

That said… there are still some differences on how a powered hub-system responds depending on how precession is blocked. The difference in response will depend on whether precession is interfered with (1) by a rigid physical obstacle, Vs (2) by a “static” force (which has a secondary precession of its own), Vs (3) by another active force from an additional motor or such. I think we will agree that the above equivalence among these 4 torques will be compromised when the interfering torque (Ff xlf) is an ACTIVE force, but not when the interfering force (Ff x lf) is a PASSIVE obstacle. (Please let me know if you disagree.)
In short, there are still a few variations to fully define depending on the nature of the different interactions involved.

I think it is clear to see that the results will change depending on whether (Ff x lf) is static (e.g. an obstacle or friction), as opposed to when (Ff x lf) is active (e.g. when a mechanical motor or method is used to apply an additional mechanical force).

Beyond common interfering torques, the interaction with centrifugal torque “Fz x lz” is unique and differs from obstacles and torques that interact with “Fp x lp” because “Fz x lz” is an outcropping from the existing torques (and therefore “Fz x lz” cannot upset the equivalence of torques mentioned above).
To say that (Fz xlz) or its deflection (Fq x lq) affects precession (Fp x lp), is tantamount to saying that (Ff x lf) interacts with (Fp x lp) more than once simultaneously. This error emerges from using ad hock methodology that has no good discipline to account for all interactions that occur with precession (there are better, more disciplined methods that are not yet accepted).

Also, it may appear trivial but should not go without saying that if an interfering torque (Ff x lf) fully blocks (Fi x li), then no interactions will occur (no precession and no hub rotation).

-
Moving on, mainly I do not agree that the centrifugal-torque (Fz x lz) can counteract the precession-torque (Fp x lp) in a way that can produce a balance-point (without introducing other factors into the mix).

I believe that you have already agreed that centrifugal-torque deflects the gyro/sphere back into the direction of hub-rotation. In other words, f(d)(Fz x lz) results in (Fq x lq)…(which moves the gyro/sphere in a direction that matches the direction of the hub-rotation).
This initial facet of the encounter between precession and centrifuge cannot produce a balance-point, and I think you already know that (but let me know if you disagree).

That brings us to the next facet of interaction, which involves your EXPECTATION that the deflected-centrifuge (Fq x lq) can have an effect, upon (Fh x lh) the hub-rotation-torque…, what that may do to (Fi x li), and subsequently how this chain of events can affect the primary precession (Fp x lp)??
I think your big question here is whether this chain of events can create a static balance-point that may then lead to propulsion?
You have previously argued that once the centrifugal-torque is deflected into (Fq x lq), then the motion of (Fq x lq) causes a “reduction” in the level of deflection from (Fi x li) into (Fp x lp).
I believe you stated that the reduced precession occurs because (Fq x lq) occurs in the same direction as hub-rotation thus reducing the effect of (Fi x li), and consequently reducing (Fp x lp) as well.

This is an error, which results from taking one’s attention away from the critical role that is played by the interfering force (Ff x lf), throughout the centrifugal interactions.
Either (Ff x lf) exists or it does not, and when (Ff x lf) is introduced, it is the sole cause of redirecting precession.
NEITHER the centrifugal torque (Fz x lz) nor its deflected motion (Fq x lq) can reduce the in/upward motion of (Fp x lp).
The centrifugal effects can only add a dimension of angular motion to the overall precession, but this added dimension of motion is ultimately caused by (Ff x lf).
When (Ff x lf) is removed, the centrifuge stops and the (Fz x lz) & (Fq x lq) torques no longer exists.

I think you also realize that the angular velocity of (Fq x lq) is slower than the angular velocity of the hub-rotation (in devices with relatively high gyro/sphere spin-rates). Therefore the velocity produced by (Fq x lq) always lags slower than the hub-rotation, unless the gyros/spheres spin at an extremely slow rate.

I am sure you know that when we gradually remove the (Ff x lf) constraint to precession (Fp x lp), hub-rotation slows-down and the resulting centrifuge becomes EXPONENTIALLY weaker. In other words, when the hub-rotation is reduced to half, the centrifugal force drops-down to one fourth etc.
Therefore completely removing the (Ff x lf) constraint makes the centrifuge dissolve rapidly if not immediately.

On the other hand, adding a DEADWEIGHT (non-spinning) mass creates hub-momentum that will continue contributing to the centrifuge (Fz x lz), while NOT contributing to the force of precession (Fp x lp) at all; this, as I think you have agreed, does provide an environment conducive to maintaining a centrifuge that can be self-sustaining as long as the torque is applied (even after the (Ff x lf) constraints are completely removed). Of course the position of this balance-point will depend on the ratios of spin to hub-rotation, and on the ratio of spinning-mass to deadweight mass.

Arguing that deadweight has a negative effect on propulsion, is a bit hallow because that is also true of all other (deadweight) components that are necessary to operate the different aspects of the device (such as the frames, arms, and many other components).

I admit that some other of my own previous arguments have emerged as hallow as well because they failed to adequately segregate (and equate) the motion-concepts of velocity Vs force & torque.
This is a mistake that often goes unnoticed because we use the word “MOTION” ambiguously to mean both acceleration and velocity, without considering that acceleration can be “Static” (in the same way that a force or torque can be “static”).
-

As I understand, your argument is that deadweight is not necessary and that the spheres mass can be static while producing propulsion (please correct me if I am wrong in the way I understand your argument).

Our reasoning tells us that a constraint must initially prevent the spheres from pivoting in/up, to allow hub-rotation and creation of centrifuge. According to EDH, once the hub-rotation achieves the correct ratio in respect to the sphere-spin, the constraints can be removed…

As I stated earlier, without some type of constraint (or without deadweight), it is not possible for continuous, self-sustained balance to exist; the hub-rotation and centrifuge fail almost as soon as the constraints are fully removed.
At a ratio of 9 to 1, the secondary precession “f(d)(Fz x lz)” derived from centrifugal-torque, is much slower than the hub-rotation that results from constraining precession (Fp x lp).

-
Regarding the balance point there is no change of mind from my part; just another case of miscommunication due to mutual misunderstandings while trying to accommodate terms to describe what we perceive.
As I have previously expressed (not too successfully), the dynamic “back-and-forth” vibrating “flutter” requires the existence of a balance point.
Please recall my analogy to a swinging pendulum where a mass moves “back-and-forth” and note that the pendulum could not do so without a balance point (a balance-point that does not move).

On our earlier discussion your statements indicated that a flutter is not necessary; this indicated to me that you perceived the spinning-mass as being static in the location of the balance point.
I am sure you are aware that a balance point and a flutter or “pulsating device” are not mutually exclusive from each other. Therefore I am puzzled why On 13 March 2008 you said:
HK* “But at this stage I cannot recognize a pulsating interaction of both spheres. In my opinion the both spheres should balance and remain at a certain position. How do you achieve the pulsating?”.

LG* I also don’t understand why you appear to reiterate that opinion in the last sentence in a previous posting of this thread.
Perhaps to remove the ambiguity in this discussion I should once again clarify that while relative location of the balance-point does not change (is “static”), the mass of the spheres does move in a sinusoidal flutter or vibration. Please let me know if you are still looking for further explanation of how the flutter is achieved.

-
Side note:
In my opinion the momentum of precession-torque does not change very much as the spheres move in/up.
However, the magnitude of centrifugal-torque INCREASES as the angle of centrifugal-attack becomes steeper.
On the other hand, the radius of the centrifugal-torque shrinks with steeper angles, and the shorter radius REDUCES the centrifuge (especially at angles closer to 90o).
-
Finally, here is another very interesting observation, which your keen mind may have already investigated.
My observation is that using a flutter (or vibration) of relatively low amplitude (e.g. 15o) and high frequency, in effect increases the proportion of time that the system spends accelerating and decelerating the mass of the sphere via precession, and via centrifuge. (In previous threads I have argued that expanding duration in acceleration zones (also “J” and beyond) is where propulsion may be harnessed.)

Most (maybe all) other devices appear to focus on the somewhat simpler PRECESSION-MOTION, as opposed to PRECESSION-ACCELERATION, “J” etc.
The reason is that most devices are designed to tap into the DIMINISHED “EQUAL-AND-OPPOSITE” reaction, displayed by the motion of precession under certain circumstances.
The more innovative design with smaller amplitude, shifts the focus toward PRECESSION-ACCELERATION (which is rather short-lived in most other common designs).
-
Regarding your comment on agreement#1, it is true that the input torque (Fg x lg) has a MAX at 0o and a MIN at 90o for gravity driven gyros. However on mechanically driven systems, the input torque (Fi x li) is determined by what the motor delivers, and that system torque is constant, in respect to the system as a whole (unless we accelerate the motor).
The centrifugal-torque can not balance with precession-torque because the centrifugal-torque always deflects precession, and the question is just a matter of degree (i.e. how fast the deflection occurs).
-
I have no problem with your modification to my agreement#5.
-
Regarding agreement#8, I think we are saying the same thing in different ways. I am also glad you now agree that a balance-point can only be achieved with very slow gyro/sphere spin rate.
-

Here are some Items we May Not Agree Upon:

Disagreement #1:
In my opinion the magnitude of the angular momentum that may be stored in the motion of precession is not always equal to the angular momentum stored in hub-rotation (this appears obvious to me).
For example, if the interfering torque (Ff x lf) is large, then the hub-rotation will get more angular velocity (even when using a sphere)…perhaps I am not understanding your statement the way you meant it (happens quite often here)… I do agree that a sphere behaves symmetrically in all planes (unless constrained by other factors), while a flywheel will not behave as fully symmetrically.

Disagreement #2:
I agree that the 100% block of precession causes the spinning object to re-deflect, effectively removing precession velocity from the normal precession plane.
However, you appear to say that the precession velocity remains unchanged… I suspect that is not what you meant to say, because when precession is fully blocked it has no velocity in the vertical plane.
Maybe you can clarify what you may have said about this?
-

All in all we are closer in the foundation basics; now we need to define the nature of the balance-point, and from there we can look into propulsion, energy… and wherever else imagination leads us.

Sorry I have not had time to find Sandy’s experiments yet (the search feature in this forum does not work).
In his experiments Sandy set the “hub-rotation” in motion to predefined velocities, and then the gyros were gradually rotated into spin, starting from very slow etc. This was repeated for different rates of “hub-rotation”.
I will continue searching for his postings (perhaps Sandy can provide his observations).
The results showed that the centrifuge is overpowered very early on. It appears that beyond some spin-threshold, no amount of hub-rotation was able to recover visible centrifuge (within the limits of the experiments conducted).

Best Regards,
Luis

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Answer: Harry K. - 15/07/2008 08:47:33
 Dear Luis,

Thank you for your detailed answer. I think we have still different opinions in some basic gyro behavior, but we are on the right way of discussion. May be I'm able to convince you of some arguments or reverse. Let's see... ;-)

I will reply to your posting later.

Regarding Sandy's link, is it this link:
http://www.gyroscopes.org/forum/questions.asp?id=768 ?

Best regards,
Harry

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Answer: Luis Gonzalez - 15/07/2008 19:04:47
 Thanks for the quick response Harry,

The thread you presented is not the one I am refering to.
The one with Sandy's experimental reporting was posted sometime in 2005.
I am looking forward to more productive interaction.

Best regards,
Luis

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Answer: Harry K. - 15/07/2008 19:49:23
 Dear Luis,

Thank you for your congratulations, however, I'm not sure if it was really helpful for better insights into gyro behavior?
This time I will not quote each passage from you posting, otherwise my posting would be too long.

First I have to mention, that the introducing of additional torques might be confusing and may block the real sight of things. I have read your posting several times and in my opinion, your way of thinking is much too complicated. Furthermore I have noted, that I still not was successful in my explanations.

It begins with your introducing of (Fq x lq) as a "secondary precession" derived from centrifuge. This torque is not necessary for better understanding. I try to explain again the gyro behavior in this context:
1. You remember that I have introduced a torque, acting against precession. This torque stands for EVERY constraint against unhindered precession movement, e.g. a counter acting torque caused by friction. I have called this torque (Ff x lf).
Precession will caused by a STATIC torque acting at the hub by e.g. an electric drive. If a counter acting torque (Ff x lf) to precession (Fp x lp) would not be present, it would not be possible for the electric drive to rotate the gyro around the hub. The electric drive would cause a STATIC torque (Fi x li), which would be deflected entirely into NON STATIC precession torque (Fp x lp). That means, the gyro would rotate in a vertical plane upwards, but the hub would remain at its current position without any horizontal rotation.

2. Now, if there is acting ANY constraint against unhindered precession movement, the gyro will deflect a part of its precession movement back to the ORIGIN, i.e. the gyro begins to rotate around the hub with a velocity, given by the amount of the counter torque against precession. That means, that the gyro rotates around the hub with a velocity derived by the counter acting torque (Ff x lf).
Because these issue are very important, once again:

A: THE CAUSE OF PRECESSION IN VERTICAL PLANE IS A STATIC TORQUE, ACTING IN HORIZONTASL PLANE AROUND THE HUB !

B: THE ROTATION AROUND THE HUB IS THE CAUSE OF ANY CONSTRAINT AGAINST UNHINDERED PRECESSION MOVEMENT !

So, if you be aware of my statements above, you know pretty well what a centrifugal force would cause in this gyro system:
1. The centrifugal force will cause a torque (Fz x lz) CCW in vertical plane against the precession torque (Fp x lp), which acts in exactly the same vertical plane but in CW direction.
2. This COUNTER torque to precession torque will cause the gyro to rotate around the hub in horizontal plane in the SAME manner, how it would do EVERY counter acting torque to precession torque! Therefore, in this case, the torque (Fz x lz) is equal to (Ff x lf), because (Fz x lz) constrains the precession movement of the gyro!
Please note that (Ff x lf) stands only for any counter acting torque against precession, nothing more.

So you can see, that the gyro behaves in a very logical sequence. Every other behavior would be not logical and therefore improbable. Anyway, this behavior covers my observations and experiences as well.

I will not go into every statement of your posting, because many of your statements are based on the above stated misunderstandings from your side.

Some more errors from you side:
1. An electric drive can be rotated in:
- Idle mode (torque against zero)
- Part load (any torque between zero and maximum)
- Full load (maximum torque)
You can prove this fact by measuring the current during operation of the electric motor!

2. It makes not any difference, if precession would be constrained by:
- Rigid physical obstacle
- A static force
- Mechanical force by a motor or anything else
- Active or passive forces
Believe me, the result will ALWAYS be a COUNTER TORQUE, which is working against precession! That's physics! ;-)

I believe this is enough for now. I hope, you will not feel offended, but I ask you to reconsider the gyro behavior under the influence of centrifugal forces.
I do not really believe in the possibility to achieve propulsion with the help of a static balance point, but I INSIST on the existence of a balance point!
This balance point can be found by using a correct design with accordant operating parameters AND by running FIRST the gyro around the hub in horizontal plane, BEFORE spinning up the gyro(s)!

I will soon start a separate thread, where I will post the accordant mathematics for calculating the balance point. In this new thread I will only discuss the mathematics, but not the theory, which should be discussed here.

Thanks, Luis, for your patience in trying to understand me... I believe you are the only one! ;-)

Best regards,
Harry


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Answer: Sandy Kidd - 16/07/2008 06:30:26
 Hello Luis,
You posted:
“In his experiments Sandy set the “hub-rotation” in motion to predefined velocities, and then the gyros were gradually rotated into spin, starting from very slow etc. This was repeated for different rates of “hub-rotation”.
I will continue searching for his postings (perhaps Sandy can provide his observations).
The results showed that the centrifuge is overpowered very early on. It appears that beyond some spin-threshold, no amount of hub-rotation was able to recover visible centrifuge (within the limits of the experiments conducted)”
.
It was my belief for some time early on in my involvement in this search, that the gyroscopic torque had to be progressively increased to overcome the centrifugal force developed in a mechanically accelerated system.
I have no doubt that this is the general belief.
Howsoever, experiment dictates otherwise.
As you correctly suggest the centrifugal force is overpowered early on, but, may I add, not so much overpowered as progressively reduced by the action of the gyroscope itself, as its rate of rotation is progressively increased.
You are also correct in your statement that no increase in rotation speed, can after a certain point reintroduce centrifugal force into the proceedings, and after that certain point, and depending on the design parameters of the system, increase in (hub) rotation speed only serves to exacerbate the situation.
Hence my disbelief in any “balance point”
This is also why I coined the term “saturation zone”
Once in this zone, dramatic changes by way of reductions to “hub” rotation speed or gyroscope rotation speed or a fair lump of both is required to recover the situation where centrifugal force is once more present in the system.
OK this is hard to swallow but this is in the world of the spinning wheel.
If experiments similar to the ones I have described are carried out, it is very easy to discern the outcome; the problem is in understanding the mechanism by which this all happens.
Sandy.


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Answer: Harry K. - 16/07/2008 09:04:22
 Dear Sandy,

Please allow me some comments to your posting targeted at Luis.

I have big respect for your person and your work in the past, but I have although my own beliefs in gyro behavior. I have tried to explain my beliefs in a logical manner but it seems that nobody can or do not want to understand it.

I have asked you several times to present more detailed data of your mentioned experiment regarding your "saturation zone", but till now I have got only these information:

- 2 opposite mounted, overhung gyros
- Gyros are fixed at an angle of 51.3° (tan 1.25) above horizontal plane
- Lever length center of gyro to center of hub: 7.5 inch
- Gyro diameter: 3 inch
- Rotation speed of gyro: 12000 rpm
- Rotation speed of hub: 325 rpm

I got these data from you in an other thread and I have asked for more details, e.g. if or how the gyros were mechanical fixed at the stated angle of 51.3 deg, however, I have not got these details. Therefore I guess that you had fixed the gyros at this defined angle in any mechanical manner.

This is a big different to my proposed design, because each gyro must unhindered move in vertical plane from 0° to 90°. If you fix the gyros at any angle defined by you, the chance to get the right angle of balance would be very small.

I have posted in the other thread, that I have calculated an angle of roughly 80° above horizontal plane, were the gyros in your experimental setup should be in balance and remain at this vertical position.
So if you do not allow the gyros to reach this position, because you have fixed them mechanically at 51.3°, an increase of hub rotation would never allow centrifuge torque to overcome the precession torque, because the relation of all parameters do not match together!
Again, the 80° position is only valid for the above stated data which you have provided. If you change any of these parameter, e.g. increasing the hub speed, an other angle of balance would be the result.

Therefore, I believe of course the rightness of your observations, but I'm sure that your device did not operate with such parameters to achieve a balance point between centrifugal torque and precession torque.
Experiments are really good, but they make no sense if they are carried out beyond useful parameters. And if you are in the dark because you have no theory, the chance to find useful parameters would be very small or impossible at all.

I have now reached a point, were I have lost the fun to continue here. If nobody read and try to understand my arguments, and if nobody can present logical arguments to reject them, it makes no sense for me to continue this discussion. Anyway, I will present my calculation formulas, maybe some visitor would be interested in. Afterwards I will quit posting here.

Regards,
Harry

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Answer: Luis Gonzalez - 16/07/2008 21:57:52
 Dear Harry,

I understand and agree with everything in your posting to me of 15/07/2008 19:49:23, up to your item “B”.
Item “B” has a small error in cause and effect; I believe that constraint causes the hub-rotation and NOT the other way round (i.e. rotation does NOTt cause the constraint). I suspect the error may have resulted from differences in language construction so I will not dwell on it.

After Item “B”, I agree with your (second) item #1., that (Fx x lz) and (Fp x lp) occur in opposite directions (as that is obvious), but (Fz x lz) is STATIC while (Fp x lp) is not STATIC (and the subtleties of that interaction may not be as obvious to all).
This leads us to your (second) item #2., which I would agree with, if I did not know that (Fz x lz) CANNOT exit without a preexisting (Ff x lf); this omission is what causes your “chicken and egg” problem.
In other words, when (Ff x fl) does not exist, then (Fz x lz) CANNOT exist either, except when the rate of gyro/sphere SPIN IS VERY LOW (do you disagree with this?).
I would say that this is a basic error from your side. Please note that when that error is corrected many of my other statements become congruent.

Perhaps even my level of complexity and detail is necessary to eliminate the ambiguity (chicken & egg) by providing the correct level of discrimination among the forces involved.
This leads us to your (third) item #2., where you expect that a system with a simple obstacle-to-precession makes use of the exact same level of torque forces as a system that introduces a new torque that forces precession to retreat through use of another secondary motor. This is definitely another one of your errors because the use of a separate torque (via a different motor) cannot be accounted without something like my equivalence equation (which you probably think is completely unnecessary).

On a different tone, I admit my error in having a limited view about torque loads on running motors, and therefore you are correct that the torque load is greater at 0o than at 90o.

Perhaps you are also correct because I think most everyone else thinks I complicate things more then necessary.
However it strikes me as odd that your response to my posting took only one day while mine took me 3 weeks and 3 days.
This means that I spent 24 times longer than you, analyzing your posting before I felt I understood it sufficiently to respond with good conscience that I had done justice to your ideas.
Therefore I also can’t help thinking that the level of complication I am addressing is not much different from yours, but just a different type.

I don’t think we should become so offended that we must give-up on the effort.
There are a few other things I want to say but I will live them for my next posting.

Best Regards,
Luis

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Answer: Luis Gonzalez - 16/07/2008 22:22:28
 Hi Harry!

Regarding your response to Sandy, you are suffering from the same affliction that attacks all who post about this subject.
Don’t be disappointed that there are not crowds of people trying to present logical arguments to your views.

You can be certain that there are a good number of eyes reading through what is developing in these postings.
We are not talking about everyday kind of science Harry. We may be pushing the limit of what is acceptable science currently and I think most people are not willing to stick their necks out that far.

The conversation we are having continues to peal back layers that thus far have only been discussed in confusing exchanges that do not even include complete sentences in many cases.
Our discussion is probably one of the most methodical ones about this subject (at this depth).
It is unfortunate that you consider my analysis too complicated and think that by introducing additional torques I am increasing the confusion and therefore blocking the clear visibility of the “real” world of spin, as you see it.

I have some observations about your technical explanations.
It looks like you expect one torque may explain most torques as a single entity, and one deflection explains all deflections as one entity. I think it’s difficult to expect others to understand and accept what you say about these monolithic deflection instances because they function in one plane at one stage and in another plane at another stage, as they respond to different forces.
In doing so you ask that we limit the language that describes the precession so that we cannot discriminate among their differences. If we adhere to this rule, then when we say precession it means the one that moves up as well as the one that moves in the same direction as the hub rotation…
Perhaps you are saying that the deflection that moves in the direction of the hub rotation should not be called precession but rather hub rotation, even though it is produced by centrifugal-torque, which could not exist without the PRE-EXISTING HUB-ROTATION?? (Chicken & Egg.)
Harry… we certainly are thinking in different ways so maybe its best if we allow each other to complete our individual explanations (no matter how complicated) and then see which one may be more plausible.

We cannot describe only a piece of the elephant and expect to achieve a complete meeting of the minds in this subject. I remember you once told me that even in a discussion with another German engineer you could not get past a certain point.

By all means Harry, present your calculation formulas, and perhaps a new visitor or one of us will understand what you want to covey.

I am going to put on my school thinking-cap and start by accepting everything you say in describing the “Elephant” until you complete your explanation.
Then after that I will only ask questions for clarification without inserting my opinion. Some of the questions may appear foolish but we have to promise to stay polite to eachother in responses.

When I am trough and have no more questions, then I will provide my opinion, and you are welcome to respond to that as well.
Does this sound like a fair plan?

Please note that while you present your side, I will probably continue to write in a separate thread about how I believe that things work…my effort will as always be to try to simplify my view of the world of spin as I see it.
There should be no reason why we should not both be able to express our points of view without interfering with each other’s ideas.

I look forward to your ideas.

Best Regards,
Luis

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Answer: Sandy Kidd - 17/07/2008 07:08:43
 Hello Harry,
You posted:-

“I have asked you several times to present more detailed data of your mentioned experiment regarding your "saturation zone", but till now I have got only these information:

- 2 opposite mounted, overhung gyros

Correct.

- Gyros are fixed at an angle of 51.3° (tan 1.25) above horizontal plane

My gyros are only suspended at that angle to allow for other manipulations.
It really does not matter what angle the gyros are at as the energy level is the same.
To carry out my manipulations I wanted to have the device operate at reasonably high angles of offset, but did not want to have to wait until the gyros got there.
The gyros once in the saturation zone will go past that angle anyway without any increase in rotation speed to the machine or the gyros. They will quite happily accelerate inwards/upwards to 90 degrees (or to a vertical position) if mechanical restrains do not get in the way.
For my purposes I did a lot of experimentation between 40 and 60 degrees of offset as it was convenient for me to do so.
As soon as a gyroscope starts its upward /inward acceleration it is in the saturation zone and displays no centrifugal force or angular momentum.
Below this level of combined rotation speeds there will be some angular momentum left to be accelerated and centrifugal force can be developed, but only if an increase in “hub” rotation speed does not carry the gyroscope into saturation. A “hub” increase in rotation speed can in fact give the same effect as an increase in gyroscope rotation speed, so that must be taken into consideration.

- Lever length center of gyro to center of hub: 7.5 inch
- Gyro diameter: 3 inch
- Rotation speed of gyro: 12000 rpm
- Rotation speed of hub: 325 rpm
ALL OK

I got these data from you in an other thread and I have asked for more details, e.g. if or how the gyros were mechanical fixed at the stated angle of 51.3 deg, however, I have not got these details. Therefore I guess that you had fixed the gyros at this defined angle in any mechanical manner.

The gyros were restrained by flat plate links with long slots which allowed screws or studs to restrain them from dropping, but allowed them to rise as the studs traversed along the slots..

This is a big different to my proposed design, because each gyro must unhindered move in vertical plane from 0° to 90°. If you fix the gyros at any angle defined by you, the chance to get the right angle of balance would be very small.

They are not fixed as such. As I previously stated in another thread in discussion and agreement with Nitro and Glenn, the triangle must never be closed.

I have posted in the other thread, that I have calculated an angle of roughly 80° above horizontal plane, were the gyros in your experimental setup should be in balance and remain at this vertical position.
So if you do not allow the gyros to reach this position, because you have fixed them mechanically at 51.3°, an increase of hub rotation would never allow centrifuge torque to overcome the precession torque, because the relation of all parameters do not match together!
Again, the 80° position is only valid for the above stated data which you have provided. If you change any of these parameter, e.g. increasing the hub speed, an other angle of balance would be the result.

My experiments without exception have shown that once a gyro begins its inward/upward acceleration it will continue quite happily to a vertical position just because there is no angular momentum in the system.
As far as I am concerned there is no balance point.

Therefore, I believe of course the rightness of your observations, but I'm sure that your device did not operate with such parameters to achieve a balance point between centrifugal torque and precession torque.

No Harry you are missing the point. The system is controlled by the action of the gyros which are capable of controlling the centrifugal force and do not have to fight it.

Experiments are really good, but they make no sense if they are carried out beyond useful parameters.

This is exactly what I think you are doing Harry

And if you are in the dark because you have no theory, the chance to find useful parameters would be very small or impossible at all.

Just because I do not agree with your line of thinking Harry does not mean I do not have a plausible theory. I have presented my findings on innumerable occasions, in simple terms as I found them, and whether anyone is prepared to believe any of it, or not, is entirely up to them. All I can say is that gyroscopes are so different in their actions which fly in the face of normal expectations. I am on a hiding to nothing trying to describe these things which must be seen to be believed, and no amount of posting to this site is going to convince any of you. Why do you really think I am sitting here 25 years after I discovered all this good stuff? Because it all sounds like figments of a deranged imagination! If I cannot convince the people on this site you will have a fairly good idea what chance I had in the other world. The primary reason I got into this site a few years ago now was to ensure my findings were not lost, well at least read. Some one sometime is going to have to gain credibility and I do not think it is going to be anytime soon.

I have now reached a point, were I have lost the fun to continue here. If nobody read and try to understand my arguments, and if nobody can present logical arguments to reject them, it makes no sense for me to continue this discussion. Anyway, I will present my calculation formulas, maybe some visitor would be interested in. Afterwards I will quit posting here.

Luis and yourself have continued on not too dissimilar lines of thought which I do not agree with, I have consequently tried to be diplomatic by shutting up and letting the pair of you get on with it.
Unless the pair of you and any others for that matter, really want to learn what spinning discs really do, you will have to build the experiment as I had to do and then and then only can we debate this stuff on equal ground..
Regards,
Sandy


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Answer: Luis Gonzalez - 17/07/2008 20:13:16
 Dear Sandy,

Thank you for your response and for confirming my recollection of your experimental results.

I agree that your reported “results” are correct (though I don’t agree with your conclusions); I have also provided reasons as to why the flywheels/spheres continue on their way up/in once deflection starts (and cannot be turned back by increasing any of the rotation parameters).

The only thing I have not openly addressed yet is why an increase in the rate of hub-rotation has the effect of increasing the rate of up/in deflection instead of increasing the resistance.

Perhaps we will have an opportunity to exchange opinions on that bit of phenomena after Harry and I are through with our exchange of presentations etc.

Best Regards,
Luis

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Answer: Harry K. - 17/07/2008 22:48:24
 Dear Luis,

Thank you for your excitation to continue with the discussion. It's very difficult to explain the own way of thinking to others and therefore misinterpretation or misunderstandings are often the result. However, let's try it again!

Here my reply to your posting from 16/07/2008 21:57:52:

Regarding item "B" you are certainly right with the error. It must be read:
"B: THE ROTATION AROUND THE HUB IS THE RESULT OF ANY CONSTRAINT AGAINST UNHINDERED PRECESSION MOVEMENT ! "

Please apologise this error in meaning.

Regarding your following comments I have to admit, that I have failed to provide deeper explanations of the basic meaning of "torque":

1. A torque can occur static OR dynamic. For example, the delivered torque of an electric motor varies according the demanded power from the work engine. So the motor may deliver different torques but at same rotation speed.
2. A static torque, e.g. the tilting torque of a free precessing gyro, does not perform physical work, because of its static appearance.
3. A dynamic torque, e.g. the deflected torque of a precessing gyro, can perform physical work, because of its dynamic appearance. However, if you try to use physical work from a precessing gyro, the gyro's precession will be deflected back to the applied tilting torque, which becomes now dynamic and the precession torque becomes static because of the applied counter acting torque to precession.

You see, there are always both manifestations of torque, static and dynamic, present if a gyro is precessing and it always an interplay of both manifestations.
- A static applied torque will cause a dynamic deflected torque.
- If precession will be blocked by a counter torque or any constraint, will cause the formerly static tilting torque to be now dynamic and the formerly dynamic precession torque will now be static.

This behavior is a fact. But important is only the fact, that the deflected torque of precession is always equal to the applied torque (Fi x li). That means that the parameters of a gyro do not have any impact to the deflected torque. It is always equal to the input torque! Different gyro parameters (spin velocity, spinning mass, material, etc.) have only an impact to the precession speed, nothing more. This issue is not trivial and it takes me long time in the past to understand this fact!
Therefore there is no need to make any differences between these both types of torque for further consideration or calculations.

Another point is my consideration of the balance point. This consideration is based on a static frame of reference in equilibrium. Please note, that a constant moved or rotated (but NOT accelerated) frame of reference is considered as "static" as well.
So the prove of a balance point has nothing to do with gyro behavior, because it is only a simple calculation of torques and forces in this frame of reference in equilibrium. This kind of calculations is my everyday job. However, the acting torques in this frame of reference in equilibrium are derived by gyro behavior and centrifugal forces.

Therefore, for better understanding, the balance theory should be divided into two parts:
1. Torques and forces caused by gyro behavior and centrifugal,
2. Consideration of these acting torques in a frame of reference in equilibrium.

If my statements above are understandable, you may answer yourself some comments of your posting:
- (Fz x lz) CAN exist without (Ff x lf), if the gyro system will be first rotated around the hub, BEFORE the gyro starts to spin. In this case (Fz x lz) is ITSELF (Ff x lf), because (Fz x lz) is the CAUSE of hub rotation! Can you agree? If not, please let me know because this issue is the most important one!
- If you apply an ADDITIONAL counter torque to precession (Ff x lf) to achieve hub rotation to cause in turn centrifugal torque (Fz x lz), then you have at the end two counter torques (Fz x lz) + (Ff x lf). For instance, this would be the case if you would fix the gyro at any defined position. In this case you have indeed found a balance point, because the gyro will always remain at its place and therefore it is in balance. ;-) Even Sandy will not deny this fact! ;-)))

However, my theory is based on free movement of the lever where the gyro is mounted, hindered in movement only by centrifugal forces. How this may be achieved have I now explained more than enough.

Regarding difference of motor drive and any other obstacles:
As stated above, I do not make any differences between kinds of constraints. A torque is torque, caused by any obstacle or a motor drive. The torque may be static or dynamic that makes no difference in gyro behavior. Also, in context with gyro behavior I see only these torques:
- Input torque (Fi x li), static or dynamic (CAUSE)
- Precession torque (Fp x lp) (EFFECT)
- Torques acting directly against (Fi x li) in same plane, e.g. friction in bearing of hub.
- Torques acting directly against (Fp x lp) in same plane, e.g. friction in bearing of lever pivot, or counter torque caused by CENTRIFUGE.

In my opinion, only this kind of torques can occur in this gyro system. Any further introduced torque may cause confusing in understanding. However, it's up to you if you prefer a more complex way of thinking. ;-)

Sorry, if I reply to your postings too fast for you, but the reason is that I have presented my theory and you are trying to understand it. Reversed, if you would present a theory and I must try to understand it, it would surely take me much longer to reply! ;-)
However, please do not think that I do not read your replies always carefully!

Now it is unfortunately too late (11:40 pm) to answer the other postings. I promise to answer them tomorrow.

Best regards,
Harry





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Answer: Harry K. - 18/07/2008 12:18:28
 Dear Luis,

Here my response to your posting from 16/07/2008 22:22:28

You wrote:
"It looks like you expect one torque may explain most torques as a single entity, and one deflection explains all deflections as one entity. I think it’s difficult to expect others to understand and accept what you say about these monolithic deflection instances because they function in one plane at one stage and in another plane at another stage, as they respond to different forces."

Yes, Luis, you are right, that I think one torque explain most torques. You should consider that in truth all mass particles of a precessing gyro move on a path around the center of spinning mass and simultaneously in 90° direction around a pivot (this pivot can match with the center of spinning mass or can be located anywhere beyond the center of spinning mass). Therefore, in truth, there is only acting one torque at a precessing gyro, the input torque (Fi x li).
Only the mechanically design of the gyro divides the single, resultant movement of each mass point into two planes, which are aligned 90° to each other.
If you be aware of this fact, it is clear that the mechanically design of the gyro allows only a division of the in truth single movement into the two planes given by the degree of freedom of the gyro design.
And it's also clear, that each mass point of the spinning mass cannot behave different in the two planes, because in truth there is acting only one movement simultaneously in both planes. Therefore angular momentum or centrifuge cannot suddenly disappear in one of the two planes.

You wrote:
"Perhaps you are saying that the deflection that moves in the direction of the hub rotation should not be called precession but rather hub rotation, even though it is produced by centrifugal-torque, which could not exist without the PRE-EXISTING HUB-ROTATION?? (Chicken & Egg.)"

We had this discussion in a previous posting. You think, centrifuge torque causes directly hub-rotation, but I think centrifuge-torque works at first against precession torque and this causes in turn hub rotation.
Actually both considerations are not important, because in both cases the result will always be hub rotation, and hub-rotation is the result of a counter torque to precession.
And as stated in my previous post, centrifugal torque can exist BEFORE precession torque, if the gyro starts to spin AFTER hub rotation. Therefore the Chicken & Egg issue is not relevant.

Luis, you can proceed with your comments, statements and the complexity of your way of thinking as you please, but we should get ahead with small steps but with steps. At the moment we only mark time! ;-)

If you do not understand some or more of my explanations, please ask again for a better explanation. I know, that I'm not good to explain something in simple and clear words. Please excuse this fault.

Best regards,
Harry


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Answer: Harry K. - 19/07/2008 13:22:16
 Hello Sandy,

Thank you for your reply. If I understand you right, the gyros are not fixed at an angle of ~51° but only suspended at this position, i.e. the gyro can move upwards in vertical position without any constraint? However, there must be a constraint, otherwise the gyros would not rotate around the hub. Do you have a plausible explanation for this fact?

You wrote:
"It really does not matter what angle the gyros are at as the energy level is the same."

I don't know what you understand by "energy level", but you can be assured that the tilting torque which is acting at the gyros will be reduced according the gyro's upwards movement and it wil be zero at the vertical 90° position. Therefore precession stops at this position.
Because of this fact you may imagine, why your experimental setup will be balanced at an angle of ~80° above horizontal plane? At this angle the precession torque is much smaller than at an angle of ~50° and thus centrifugal forces will have more influence to the system.
Now you may argue that centrifugal forces are reduced as well because the rotation radiius in horizontal plane is also reduced? That's of course right but the vertical lever arm (lz) of the acting centrifugal force generates a bigger centrifugal torque (Fz x lz) which works against the now smaller precession torque (Fp x lp).
Please note that that if the gyros would rotate in horizontal (0°) plane, centrifugal torque would be zero although centrifugal force (Fz) would be maximum, because the lever arm (lz) is zero at this position!
Therefore if the gyro starts to spin after established hub rotation, the gyro would immediately begin to move upwards, regardless how slow the gyros would spin! But during the upward moving (precessing) of the spinning gyros, precession torque would become smaller and centrifuge torque would become bigger, however assumed that hub rotation speed and spinning speed of the gyros would not be changed.

Believe me or not, I know that you can only be convinced by providing a mechanical prove. Forgive me, I'm a simple design engineer and have ever designed machines first in theory before they were afterwards built. Most of these machines are still working! ;.) However, I will think about a low budget test setup...

Anyway, I would be very interested to read about your "plausible theory" of a saturation zone. All what I have read by now has unfortunately nothing to do with physics or logic. Maybe you can post a link?

Regards,
Harry

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Answer: Luis Gonzalez - 20/07/2008 04:11:00
 Hi Harry,
I hope you are having a nice weekend.
Your postings tell me we are headed in the right direction (though I don't agree with everything).

It’s been a busy weekend with not enough time for gyro-stuff.
My next posting will focus on the balance point, and I will avoid all other complicating subjects, except where thy play a part in the intricacy of the balance point.

I will post soon, best Regards,
Luis

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Answer: Luis Gonzalez - 23/07/2008 23:23:10
 Dear Harry,

Re: Your posting of 17/07/2008 22:48:24
I will avoid all unresolved issues that are not directly relevant to the “balance-point” of the “gyro-hub” system.
I will also introduce only complexities that are necessary to explain the point more thoroughly (even if these may require some effort). Oversimplification of the complex leads to error and misunderstandings.
At this stage, my sole focus are; (a) the “balance-point” in the context of the “gyro-hub” system, (b) how you truly perceive this balance-point occurs, and (c) most important whether you have explained the balance-point correctly and clearly enough.

Your statements appear to indicate that you believe your explanation of a “balance-point” between (Fp x lp) and (Fz x lz), in a “gyro-hub” system, has been clearly and completely explained.
I am sorry to say that you have not fully connected all the dots yet.

Your explanation has succeeded in stating a number of basic items, some which are obvious and others that we have agreed about (and this is a good thing).
However, you have failed to bring these (somewhat disjointed or not fully connected) facts into a cohesive explanation; your explanations do not take into account important interdependencies among these factors, within the environment of the “gyro-hub” system (perhaps a result oversimplification, or something else).
Interdependencies generally increase the level of complexity (unfortunate, but maybe necessary).

Please note that I will only address the third quarter of your 17/07/2008 22:48:24 posting. I will ignore the first half because I agree with 98% of it, and will ignore the last quarter because I agree it covers a subject that we don’t need to discuss yet or here.

Below are your statements (from the 3rd quarter of your posting), about which I need further clarification on how they hold together for a coherent explanation of your balance-point between (Fp x lp) and (Fz x lz) within a gyro-system.

Harry Said:
#HK Start#
(A)
“…for better understanding, the balance theory should be divided into two parts:
1. Torques and forces caused by gyro behavior and centrifugal,
2. Consideration of these acting torques in a frame of reference in equilibrium.
(B)
If my statements above are understandable, you may answer yourself some comments of your posting:
- (Fz x lz) CAN exist without (Ff x lf), if the gyro system will be first rotated around the hub, BEFORE the gyro starts to spin. In this case (Fz x lz) is ITSELF (Ff x lf), because (Fz x lz) is the CAUSE of hub rotation! Can you agree? If not, please let me know because this issue is the most important one!
(C)
- If you apply an ADDITIONAL counter torque to precession (Ff x lf) to achieve hub rotation to cause in turn centrifugal torque (Fz x lz), then you have at the end two counter torques (Fz x lz) + (Ff x lf). For instance, this would be the case if you would fix the gyro at any defined position. In this case you have indeed found a balance point, because the gyro will always remain at its place and therefore it is in balance. ;-) Even Sandy will not deny this fact! ;-)))
#HK End#

***
Here is my take on your explanation above:
A) I agree on the need to consider and view from perspectives of frames-of-reference but we must take into account all forces that exist WITHIN the frame, no matter how their causes originate.

B) I agree that when the gyros are NOT spinning, (Fz x lz) exists without need for (Ff x lf); however “Fz x lz) IS NOT the CAUSE of hub-rotation (this is an error in your explanation… a stretch of the imagination).
We know that when the gyros are NOT spinning (Fi x li) is the sole cause of hub-rotation, and (Fz x lz) is the result from hub-rotation rather than causing it (wouldn’t you agree?).
Therefore I DISAGREE with your statement in this regard… but as you suggested, I DO AGREE that this is a most important issue.

I think “(Ff x lf)” is causing your confusion because, when the gyros are NOT spinning, “(Ff x lf)” is not necessary to perform calculations or ANYTHING else (so why introduce it?). When gyros are not spinning, there is no (Fp x lp) that needs to be blocked in order to obtain hub-rotation; the hub-rotation occurs without need for (Ff x lf).)
Please clarify this error.

C) I agree that the centrifuge that results from fixing the pivot-arms of spinning gyros (at an angle between 0o and 90o) produces (Fz x lz) in addition to the necessary (Ff x lf), but you are missing some very important points.
This important points are that the gyros are spinning, and that the magnitude of (Ff x lf) is one of the factors that determines the magnitude of (Fz x lz); therefore extricating (Ff x lf) out of the system simultaneously removes (Fz x lz), and as a result (Fp x lp) takes over rapidly. (You see, this does not mean that the centrifuge “disappears”; it means that hub-rotation stops without (Ff x lf), and therefore the centrifuge stops too.)
It does not take much to see that (Fz x lz) depends on hub-rotation, and hub-rotation depends on (Ff x lf) when the gyros ARE SPINNING. (Naturally when the gyros are NOT spinning (Ff x lf) is not necessary for calculation of (Fz x lz), as explained previously.)

D) This brings us to your last 2 sentences that follow immediately after your above statements, where you say:
#HK Start#
“However, my theory is based on free movement of the lever where the gyro is mounted, hindered in movement only by centrifugal forces. How this may be achieved have I now explained more than enough.”
#HK End#

***
My response is that again I am sorry but the cohesive answer that I think you are trying to provide requires further explanation.

You have basically stated:
A) Consider toques & forces of gyro behavior in a frame of reference in equilibrium.
B) (Fz x lz) exist without (Ff x lf) when gyros are NOT spinning.
C) When gyros are spinning and (Ff x lf) is introduced, then (Ff x lf) + (Fz x lz) interfere with (Fp x lp).
D) Your conclusion: When the pivot levers are free to move then only (Fz x lz) will interfere with (Fp x lp).

I am afraid your conclusion does not follow logically from the premises present.
Perhaps you need to be more exact about the frames of reference; and why an occurrence that requires a condition of “NO-SPIN”, should be extended into a situation where SPIN EXISTS, in order to prove your theory?
The burden of proof is on the one explaining their theory.

Harry, I will post an answer to your posting of 18/07/2008 12:18:28 as soon as I find some time.

Best Regards,
Luis


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Answer: Luis Gonzalez - 25/07/2008 20:27:48
 Dear Harry,
As I understand it, you say the magnitude of precession-torque (Fp x lp) is only affected by the input-torque (Fi x li), and NOT affected at all by the rate if spin of the gyros (please correct me if I am wrong).

Can you please explain why very slow-spinning gyros would have smaller angle of equilibrium than gyros that spin a bit faster? (Equilibrium being the balance-point between centrifuge and precession torques.)
In other words, why does the equilibrium angle of the leavers increase as the spin-rate of the gyro is gradually increased (as the hub-rotation is held constant)?

On the surface it looks like faster gyro-spin increases precession-torque (Fp x lp), but we know that is not the real cause because we know (Fp x lp) is only affected by (Fi x li) and not at all by the gyro’s spin rate.

I am eager to read your analysis and response regarding this interaction.

P. S.
I am also curious what happens to the tilting force (Fi) between -90o and 0o (i.e. between 270o and 0o)?
(Since we feel sure that (Fi) is reduced as the gyro moves from 0o to 90o)
I will wait for your responses before responding to your posting of 18/07/2008 12:18:28.

Best Regards,
Luis

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Answer: Luis Gonzalez - 30/07/2008 22:53:42
 Dear Harry,
I figured out the answer to my last posting, but the answer came along with a few more questions and observations.

When the gyros/spheres are not spinning and the hub has settled into a regular rate of rotation, the hub-motor’s torque-load is virtually zero.

When the spheres begin to slowly spin in gradual minute increments, the torque load of the hub-motor also increases by very small amounts.

No matter how fast the hub turns, the actual engaged-torque depends on both the spin-rate and the hub’s rotation-rate; the relative frequency of these 2 angular velocities determine the true value of (Fi x li).
As the spin creates a load, the hub-spin cannot exceed the torque that the hub-motor can deliver (if the hub-motor does not have the torque-capability it cannot achieve the appropriate rotation rate against the spin-rate).

Even a very slow hub-rotation can yield a very large (Fi x li), if the sphere’s spin-rate is sufficient, and the hub-motor delivers a sufficiently robust torque.

I think it can be confusing to take a narrow view, from a perspective that relies completely on traditional mathematical notation, because there are multiple interactions occurring and evolving simultaneously over a timeframe that can vary depending on the ratios of hub-rotation to spin-rate.
Simple answers can make us go in circles as we keep returning with new questions to incomplete stand-alone answers that at times may appear to contradict each other (depending on the dynamics and the ratios).

To say that (Fi x li) is the only source of torque, may be factual but it is a gross simplification of the truth.
When one truth is stated alone, it can camouflages a number of other facts that are just as important to the full understanding of the interactions and to an appropriate theory.

It may be more accurate to say that (Fp x lp) is affected by the effective ratio between the applied-torque and the spin-rate of the sphere.

As other gyro enthusiasts, I have also wondered about the exact interactions that take place when the sphere-spin is very slow and a large torque is applied, or when a fast rotating-hub exists, followed by a very slow spinning of the spheres.

The equation for precession forecasts a fast velocity of precession when spin-rates are slower.
Just how fast can precession accelerate given a sufficiently robust torque-source combined with very slowly spinning spheres?

Best Regards,
Luis

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Answer: Harry K. - 31/07/2008 07:54:01
 Dear Luis,

I will answer your postings at the weekend. At the moment I have less time. One note to your last posting. I believe your way of thinking goes in the right direction. For better understanding why (Fi x li) is the only source of torque, you could imagine a gyro as a gearbox, which adjusts an input torque to a certain input velocity according to the gyro parameters (spin rate, spinning mass, geometric shape, etc.).
Also you should imagine, that hub rotation only can occur, if there is a resistance in precession plane against precession torque (Fp x lp). If this resistance is equal to the input torque (Fp x lp = Fi x li), then you can do all necessary calculations because you have a given torque (Fp x lp = Fi x li) and a given hub rotation velocity ( = precession velocity caused by Fi x li). This way of thinking is indeed a simplification but I don't see any mistakes in this way of thinking.
A balance point simply achieves a resistance to precession in form of (Fi x li) = (Fp x lp) and this circumstance causes in turn the hub rotation.
This may be confusing but it is clear logical thinking.

But more details at the weekend.

Best regards,
Harry

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Answer: Luis Gonzalez - 02/08/2008 16:01:17
 Dear Harry;
I have to say something regarding your posting of 18/07/2008 12:18:28

I am surprised you said that I think centrifuge CAUSES hub-rotation in any form or way.
You said:
#HK# “You think, centrifuge torque causes directly hub-rotation, but I think centrifuge-torque works at first against precession torque and this causes in turn hub rotation.” #HK#

I am sure you couldn’t have meant this, as I’ve labored to make it clear that hub-rotation is the causes of centrifuge (not the other way round); it is so basic.
Please note I have always stated that (Fq x lq) = f(d)(Fz x lz), is SLOWER than hub-rotation.

Please let’s be sure to quote each other correctly.
Best Regards,
Luis

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Answer: Harry K. - 03/08/2008 11:24:07
 Hello Luis,

Regarding your last posting from 02/08/2008 16:01:17:

Sorry, if I have misinterprted your beliefs. Due to your comments I was sure you would think so. And by the way, I didn't quoted you in this context, but only stated my belief how you may think in this context.
Here an example of your posting from 14/07/2008 22:26:13 :

"I want to introduce (Fq x lq) = f(d)(Fz x lz) as a secondary precession derived from centrifuge."

This "secondary precession" occurs in the same plane as hub rotation and therefore I thought you would think that centrifuge causes hub rotation.


You wrote:
"Please note I have always stated that (Fq x lq) = f(d)(Fz x lz), is SLOWER than hub-rotation. "

...And this exactly this statement is in my opinion wrong! The hub rotation speed will be accordant to the torque caused ba centrifuge. The introduction of (Fq x lq) is not helpful and not necessary at all. Therefore I will not use this term in my further explanations.

Best regards,

Harry

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Answer: Luis Gonzalez - 03/08/2008 17:34:06
 Dear Harry,

Thank you for your response.
I just needed to clarify my opinion.
In clarifying the misinterpretation we have a better view of our disagreement.

I say (Fq x lq) = f(d)(Fz x lz), is SLOWER than hub-rotation.
You say that this statement of mine is wrong.
You said:
#HK#
“The hub rotation speed will be accordant to the torque caused by centrifuge.”
#HK”

This is the core of our current disagreement.
It explains that you believe centrifugal-torque is sufficient to overcome precession-torque at a SPECIFIC ANGLE determined by the strength of the centrifugal-torque (Fz x lz) as opposed to the strength of precession-torque (Fp x lp), as you have stated many times.

Here is what I can put together about your view of the 2 opposing torques, based on your previous statements:
# (Fp x lp) = (Fi x li) and (Fi x li) is determined by the spin-rate, spinning-mass, and geometric shape etc (do you need to expand this side of the equation?).
# (Fz x lz) =???... This side of the equation needs your explanation.

I do not know your explanation about what gives rise to the existence of (Fz x lz); where does (Fz) come from in your model; what is the origin cause of (Fz)?

In my model (Fz) results from (Ff); what about yours?
In my model, (Ff x lf) results from non-spinning mass (“deadweight”); what about yours?

I know that hub-rotation occurs when gyro/spheres spin very slowly; I have a long and complex explanation for that phenomenon, which I think you will not appreciate at this point, so I will not present it now.

For now the moment I am interested in understanding your view of the balance-point, and if you think hub- rotation results from (Fz x lz), then where does (Fz x lz) come from?

Best Regards,
Luis

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Answer: Harry K. - 03/08/2008 21:43:03
 Dear Luis,

Thank you also for your quick reply! It's really hard to convince you about my point of view regarding gyro behavior. Please find my comments below:

LG wrote:
"It explains that you believe centrifugal-torque is sufficient to overcome precession-torque at a SPECIFIC ANGLE determined by the strength of the centrifugal-torque (Fz x lz) as opposed to the strength of precession-torque (Fp x lp), as you have stated many times. "

Yes of course, THAT IS my belief. I have tried to explain this belief from the very beginning!

LG wrote:
"# (Fp x lp) = (Fi x li) and (Fi x li) is determined by the spin-rate, spinning-mass, and geometric shape etc (do you need to expand this side of the equation?). "

Exactly correct! But please note, that at this point there is NO hub-rotation! The STATIC torque (Fi x li) in horzontal plane causes a DYNAMIC torque (Fp x lp) in vertical plane!
Do you agree with this fact or do you need more explanation?

LG wrote:
"# (Fz x lz) =???... This side of the equation needs your explanation.

I do not know your explanation about what gives rise to the existence of (Fz x lz); where does (Fz) come from in your model; what is the origin cause of (Fz)?
"

The torque (Fz x lz) is caused by hub rotation and occurs if ANY mass will be rotated around a pivot. To achieve that this torque will act, the gyro must not spin during hub rotation. The hub rotation is caused by (Fi x li) and (Fi x li) behaves now as a dynamic torque because the gyro is not spinning and therefore acting like usual dead weight.

If in a second step the gyro begins to spin, tie torque (Fi x li) will now be immediately deflected to precession torque (Fp x lp). But because the torque (Fz x lz) is still acting because it was acting BEFORE the gyro began to spin, the centrifugal torque (Fp x lp) acts against precession torque (Fp x lp) and causes IN TURN again hub rotation!

In other words, centrifugal torque (Fz x lz) transfers a part or the complete precession torque to hub rotation, i.e. STATIC input torque will become DYNAMIC because of the PRESENCE of centrifugal torque. If the COMPLETE precession torque (Fp x lp) will be transferred to hub rotation, the gyro system will be in balance, because in this case is (Fz x lz) equal to (Fp x lp) and in turn equal to (Fi x li).

Do you understand now what I have tried to explain since many, many postings? - I don't know how to explain more clearly, sorry.

LG wrote:
"In my model (Fz) results from (Ff); what about yours?
In my model, (Ff x lf) results from non-spinning mass (“deadweight”); what about yours?"

As stated many times before: Fz is a kind of Ff, because Ff stands for ANY force acting against precession torque, e.g. friction, mass inertia, CENTRIFUGAL, etc!
Thus (Ff x lf) stands for general counter torque to precession torque and includes centrifugal torque (Fz x lz) as well as any other counter torque, which may occur in this context.
I hate myself that I have introduced (Ff x lf) because it was apparently not helpful for explanation in any way!

LG wrote:
"I know that hub-rotation occurs when gyro/spheres spin very slowly; I have a long and complex explanation for that phenomenon, which I think you will not appreciate at this point, so I will not present it now."

Keep in mind that hub rotation occurs if there is ANY CONSTRAINT to precession! So it doesn't matter slow a gyro spins! If you do not understand this basic fact, you will never understand my balance theory.

LG wrote:
"For now the moment I am interested in understanding your view of the balance-point, and if you think hub- rotation results from (Fz x lz), then where does (Fz x lz) come from?"

I have again answered this question above. I would be very happy if you could agree at last. I get tired...

Best regards,
Harry

P.S. Should I answer also your other both postings or is this answer sufficient for you?


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Answer: Harry K. - 03/08/2008 21:54:39
 Hello Luis,

Please ignore my last posting! There happened some typos. Here is the corrected version:

Dear Luis,

Thank you also for your quick reply! It's really hard to convince you about my point of view regarding gyro behavior. Please find my comments below:

LG wrote:
"It explains that you believe centrifugal-torque is sufficient to overcome precession-torque at a SPECIFIC ANGLE determined by the strength of the centrifugal-torque (Fz x lz) as opposed to the strength of precession-torque (Fp x lp), as you have stated many times. "

Yes of course, THAT IS my belief. I have tried to explain this belief from the very beginning!

LG wrote:
"# (Fp x lp) = (Fi x li) and (Fi x li) is determined by the spin-rate, spinning-mass, and geometric shape etc (do you need to expand this side of the equation?). "

Exactly correct! But please note, that at this point there is NO hub-rotation! The STATIC torque (Fi x li) in horizontal plane causes a DYNAMIC torque (Fp x lp) in vertical plane!
Do you agree with this fact or do you need more explanation?

LG wrote:
"# (Fz x lz) =???... This side of the equation needs your explanation.

I do not know your explanation about what gives rise to the existence of (Fz x lz); where does (Fz) come from in your model; what is the origin cause of (Fz)?
"

The torque (Fz x lz) is caused by hub rotation and occurs if ANY mass will be rotated around a pivot. To achieve that this torque will act, the gyro first must not spin during hub rotation. The hub rotation is caused by (Fi x li) and (Fi x li) behaves now as a dynamic torque because the gyro is not spinning and therefore acting like usual dead weight.

If in a second step the gyro begins to spin, the torque (Fi x li) will now immediately be deflected to precession torque (Fp x lp). But because the torque (Fz x lz) is still acting, because it was acting BEFORE the gyro began to spin, the centrifugal torque (Fp x lp) acts against precession torque (Fp x lp) and causes IN TURN again hub rotation!

In other words, centrifugal torque (Fz x lz) transfers a part or the complete precession torque to hub rotation, i.e. STATIC input torque will become DYNAMIC because of the PRESENCE of centrifugal torque. If the COMPLETE precession torque (Fp x lp) will be transferred to hub rotation, the gyro system will be in balance, because in this case (Fz x lz) is equal to (Fp x lp) and in turn equal to (Fi x li).

Do you understand now what I have tried to explain since many, many postings? - I don't know how to explain more clearly, sorry.

LG wrote:
"In my model (Fz) results from (Ff); what about yours?
In my model, (Ff x lf) results from non-spinning mass (“deadweight”); what about yours?"

As stated many times before: Fz is a kind of Ff, because Ff stands for ANY force acting against precession torque, e.g. friction, mass inertia, CENTRIFUGAL, etc!
Thus (Ff x lf) stands for general counter torque to precession torque and includes centrifugal torque (Fz x lz) as well as any other counter torque, which may occur in this context.
I hate myself that I have introduced (Ff x lf) because it was apparently not helpful for explanation in any way!

LG wrote:
"I know that hub-rotation occurs when gyro/spheres spin very slowly; I have a long and complex explanation for that phenomenon, which I think you will not appreciate at this point, so I will not present it now."

Keep in mind that hub rotation occurs, if there is ANY CONSTRAINT to precession! So it doesn't matter how slow a gyro spins! If you do not understand this basic fact, you will never understand my balance theory.

LG wrote:
"For now the moment I am interested in understanding your view of the balance-point, and if you think hub- rotation results from (Fz x lz), then where does (Fz x lz) come from?"

I have again answered this question above in this posting. I would be very happy, if you could agree at last. I get tired...

Best regards,
Harry

P.S. Should I answer also your other both postings or is this answer sufficient for you?


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Answer: Luis Gonzalez - 11/08/2008 18:54:11
 Dear Harry,

Thank you for your answers and please accept my apology for the slow response. There is so much to do in the summer, and in an Olympics year there is even less time left.

I will post a more complete answer next weekend.
For now, I have been analyzing your perception of the balance-point from your simplified perspective and find the mechanical interactions (as you describe them) to be consistent with my model of spin dynamics (please rest assured I understand your statements even if the English requires a little effort sometimes). Also, thank you for introducing notation that facilitates easier conversation as well as deeper analysis.

One of the errors that I see in your static balance calculation (the error that I will present in this and my next posting) involves the effects of the elevation angle, on the two torques involved.

Our point of non-agreement is about the MAGNITUDES of the 2 torque involved; NOT any difference in the way we perceive the dynamics.
In other words, it’s all about the CALCULATIONS.

I) In short, the angle (from 0o to 90o) affects the precession-torque (Fp x lp) in a different way than it affects centrifugal-torque (Fz x lz). We both know that.

II) The effect of the elevation angle is such that (Fz x lz) CANNOT be larger than (Fp x lp); and only at exactly 0o and 90o do both torques become equal to zero (under the configuration we are discussing). This is where you disagree! (Note – very slow spin includes other more complex factors that we can discuss if you wish.)

III) The reason for my statement is that:
(1) - (Fp x lp) is affected by a factor of “Cos(A)”, but…
(2) - (Fz x lz) is affected by a factor of “Sin(A) multiplied by Cos(A)”…where “(A)” is a given angle between 0o and 90o.

As you are familiar, “Cos(A)” starts with a value of ONE (at 0o), and decreases to ZERO (at 90o).

On the other hand, “Sin(A) x Cos(A)” starts with a value of ZERO (at 0o) and increases to 0.5 at 45o, and then decreases back to ZERO again (at 90o).
This is where the calculation ERROR exists because even at 80o (Fz x lz) is still smaller than (Fp x lp).
(I am aware there are other variables and we can discuss their effect too but it’s a bit more complex).

The two curves “Cos(A) & Sin(A) x Cos(A)” never intersect, which means that the factor of precession-torque is always larger than the factor of centrifugal-torque.
Anyone can plot graphs of the 2 curves to verify that my statements are true.
And, if anyone is curious why these 2 curves represent the different factors of precession-torque and centrifugal-torque, please let me know.

I will present my reasoning next week so that all readers will understand why I have introduced this comparison.

Best Regards,
Luis

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Answer: Luis Gonzalez - 16/08/2008 14:11:21
 Dear Harry,

In my last posting I stated that precession-torque (Fp x lp) is affected by a factor of Cos(A), and that centrifugal-torque is affected by a factor of Sin(A) x Cos(A).

To those who have not found the reason for these seemingly obtuse statements, here is the reason I introduced these 2 factors:

The “Cos(A)” times the length of the hub-radius (pivoting arms) adjusts for the shrinking rotation radius; because as precession causes the arms to raise, the circles formed by the rotating mass become smaller (at 90o the length of the hub rotation radius is zero).
(Those with Trig knowledge understand, others can lookup table values in the web).

The “Sin(A) times the length of the hub-radius (pivoting arms) adjusts for the increasing fulcrum leverage that occurs as horizontal centrifugal-force applies torque using the mass at the end of the hub’s pivot arms.

Therefore as precession causes the arms to rise, we have a shrinking circumference of the hub perimeter’s rotation, and an increasing leverage on centrifugal-torque.

- Centrifugal-Force equals Mass x Velocity Squared, over rotation Radius etc; M x (V**2) / (R x Cos(A)).
- And, V**2 equals (2(pi)R x Cos(A) x C/t)**2; where C/t is the cycles per time unit.
- Centrifugal-torque requires multiplication by (R x Sin(A)) for the leverage length at a given angle.
- The calculation of centrifugal-torque is M x (R x Sin(A) / R x Cos(A)) x (2(pi)R x Cos(A) x C/t)**2
- Cancelling all Radii pairs, centrifugal-torque equals M x (R x Sin(A) x R x Cos(A)) x (2(pi) x C/t)**2
- The R x Sin(A) equals (lz); and the R x Cos(A) equals the rotation Radius.
- Thus (Fz x lz) equals Mass times Sin(A)Cos(A) times (2(pi) x R x C/t) Squared.
The important part is the “Sin(A)Cos(A)” factor.

- Precession-torque is equal to the input torque (Fi x li), where the factor (li) is the rotation Radius of the hub-perimeter, which shrinks as the arms rise.
- Therefore (li) equals R x Cos(A).
- And (Fi x li) equals (Fi x R x Cos(A)), which equals (Fp x lp).
The important part is the “Cos(A)” factor.

This should justify the use of the 2 parameters I introduces in my last posting.
Harry, please let me know if you disagree with the use of the 2 parameters or their effect as presented in my previous posting. For example:
Q#1) Do you disagree that “Cos(A)” remains greater than “Sin(A) x Cos(A)” from 0o to 90o?
Q#2) Do you disagree that that Cos(A) is the factor of adjustment for (Fp x lp) as the angle changes from 0o to 90o?
Q#3) Do you disagree that that Sin(A) x Cos(A) is the factor of adjustment for (Fz x lz) as the angle changes from 0o to 90o?

I will wait for your response before delving into any deeper analysis of the OTHER factors that may be expected to produce the “static balance-point”.
I will also wait before addressing the more complex behavior of slow-spinning gyro/spheres.

I want to start addressing the question of “PROPULSION”; therefore I will start another thread to begin a discussion that addresses the following questions:
1) Why someone might expect a static balance-point to produce propulsion.
2) Why I believe that a static balance-point CANNOT produce propulsion.
3) How I expect that dynamic gyro/sphere motions (surrounding a special kind of balance-point) should produce propulsion and lift.

I am looking forward to your initial answers to my introduction of the 2 parameters.

Best regards,
Luis

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Answer: Harry K. - 16/08/2008 15:11:34
 Dear Luis,

Please also excuse my late response for the same reasons than yours (Olympic games, summer...). :-)

Maybe we have a different way of calculating. Let's see who is right and who is wrong.

Re I) Yes of course, Luis!

Re II) I had never claimed that (Fp x lp) may become larger than (Fp x lp)!? This can only occur if Fp would become zero, i.e. if the gyro would not spin.
I only claim that (Fz x lz) become equal to (Fp x lp) (-> BALANCE!), nothing else!

Re III) Unfortunately you are partly wrong:
Re (1) - (Fp x lp) is affected by a factor of “Cos(A)”, but...
Re (2) - (Fz x lz) is affected by factor of “Sin(A)” AND factor of “Cos(A)”:
- Fz is affected by factor “Cos(A)”
- lz is affected by factor “Sin(A)”

That's a significant difference to your belief because if both torques will be in relationship, the factor "Cos(A)” will be cancelled! The factor "Cos(A)” remains at the end and is necessarry as well to calculate the angle of balance by "ArcSin(A)".

You wrote:
"As you are familiar, “Cos(A)” starts with a value of ONE (at 0o), and decreases to ZERO (at 90o)."

For this statement there is no need to quote trigonometric formulas. It is sufficient to think logical:
- at 0 deg. the radius is not shortened because it is aligned in horizontal plane
- at 90 deg. the radius is zero because it is aligned in vertical plane

You wrote:
"On the other hand, “Sin(A) x Cos(A)” starts with a value of ZERO (at 0o) and increases to 0.5 at 45o, and then decreases back to ZERO again (at 90o).
This is where the calculation ERROR exists because even at 80o (Fz x lz) is still smaller than (Fp x lp).
(I am aware there are other variables and we can discuss their effect too but it’s a bit more complex).

The two curves “Cos(A) & Sin(A) x Cos(A)” never intersect, which means that the factor of precession-torque is always larger than the factor of centrifugal-torque.
Anyone can plot graphs of the 2 curves to verify that my statements are true.
And, if anyone is curious why these 2 curves represent the different factors of precession-torque and centrifugal-torque, please let me know."

No your conclusion is wrong because of the issue that "Cos(A)“ will be cancelled as explained above.

You wrote:
"I will present my reasoning next week so that all readers will understand why I have introduced this comparison. "

I could present all relationships and necessary equations that all readers, who are able to calculate moderate mathematics, could understand everything. However, I will wait because I can still not believe that it is so difficult to find the right way... ;-)

Best regards,
Harry


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Answer: Harry K. - 16/08/2008 15:15:32
 Dear Luis,

My previous posting belongs to your posting from 11/08/2008 18:54:11. I didn't noticed that you wrote another posting in the meanwhile!

Best regards,
Harry

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Answer: Harry K. - 17/08/2008 00:51:56
 Dear Luis,

Here is my answer to your last posting. I think I have already answered your questions in my previous posting but I want to give some comments regarding your calculations.
In my opinion, you make your equations needless complex by using cirumferencial velocity and rotatonal speed instead of using directly radius and angular velocity. If I remember correctly I have sent you my gyro related formulary via emai. If you do not find it, I can send it again. Please let me know.

Here are my equations:
1. Centrifugal force (Fz)
Fz = mG x R x cos(a) x Wp**2

[mG... dead weight mass of gyro]
[(a)... angle alpha]
[R... radius of lever arm]
[Wp... angular velocity of precession ( = angular velocity of input torque)]

2. Centrifugal torque (Tz)
Tz = Fz x lz <=> (mG x R x cos(a) x Wp**2) x lz
lz = R x sin(a)
Tz = mG x R x cos(a) x Wp**2) x R x sin(a) <=> mG x cos(a) x sin(a) R**2 x Wp**2

3. Precession torque (Tp)
Tp = Fp x lp
Precession torque is equal to input torque (Fi x li)
li = R x cos(a)
Tp = Fi x R x cos(a)

You see that the equations have become much clearer and less complex.
The important thing for the next step of calculation is to make a relationship between precession torque, gyro dead weight, gyro spinning mass and angular velocity of spinning gyro. If that is done, you are able to equate centrifugal torque and precession torque. At this step you will see, that cos(a) will be cancelled and thus your argument regarding the 2 factors of sin(a) x cos(a) at centrifugal and cos(a) at precession is fallen.
Therefore my answers to your questions are as follows:

Q# 1) Yes I disagree because cos(a) will be cancelled and thus sin(a) is the only remaining factor.
Q# 2) No I don't disagree because I have never claimed anything else.
Q# 3) No I do not disagree that sin(a) x cos(a) is the factor of adjustment for (Fz x lz), however, you have not cnsidered the equation of the relationship between centrifugal torque and precession torque. You must consider this relationship, because of the condition of equilibrium!

To be honest, but if I were you, I would wait to start a new thread until I have understood all the basics of the theory stated in this thread.
Please let me know if you need additional help with further steps or equations.

Best regards,
Harry


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Answer: Luis Gonzalez - 20/08/2008 21:57:44
 Dear Harry,
I just noticed you responded quickly.
Will probably get to read it this weekend (lots of new projects at work).

Best Regards,
Luis

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Answer: Luis Gonzalez - 23/08/2008 19:06:21
 Dear Harry,

I had a little time to read your posting of 16/08/2008 15:11:34 and am a bit disappointed at the typing error, which compromises the meaning and significance of your statements. However the most disappointing, are the cavalier errors in logic (as well as misrepresentation of my statements).

Perhaps we can prevent lengthy misunderstandings by just correct typing errors before we post.

Best Regards,
Luis

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Answer: Luis Gonzalez - 23/08/2008 19:45:18
 Dear Harry,

I see that on your subsequent posting of 17/08/2008 00:51:56 you corrected the glaring typing error in your previous posting (though the correction was not mentioned).

This entire thread is about your theory but you have failed to prove it and have only presented snippets that would require others to embrace your theory by faith only.

I have said that I delve in matters of spin as a hobby (just for fun), and you are a claimed professional in the field (rare in this forum).
Yet I introduce equations and factual concepts that are fresh to this forum (and to this thread).
When I introduce logical insights (that cannot be disputed) you dismiss them as known facts, even though I have never seen them mentioned in this forum (I guess anything deduced logically from facts may be presumed to be already known).

It seems unusual to only addressed equations and concepts only in response to what I (and others) have stated, when you should be introducing concepts connected by equations, in support of your innovative theory on static balance (instead of only responding)?

The word “WRONG” rolls-off with great ease in your recent responses, however the equations, in your last posting of 17/08/2008 00:51:56, confirm that my statements are correct.
(This has happened before in previous disagreements and we called them misunderstandings.)
The language-barrier may be hindering both the ability to understanding and the ability to express ideas clearly, and this appears to include the correct use of the word “WRONG”.

I have provide significant indication that precession-torque is always greater than the centrifugal-torque, at all angles between 0o and 90o. Your posted equations have confirmed it as much as mine did (though in simpler notation), and the facts are recorded in this thread for anyone willing to do the high-school math (and able to discern through the postings).

You confirmed my equations and thus proved me correct (though you claimed I was wrong), but have not yet proven your theory. I am sorry to have to say this but it is true.
-

I am also afraid that your answer to Q#1 CANNOT be correct!
In Q#1 I asked:
-Do you disagree that “Cos(A)” remains greater than “Sin(A) x Cos(A)” from 0o to 90o?-
Your answer to Q#1 was:
#Harry’s answer was ## “Yes I disagree because cos(a) will be cancelled and thus sin(a) is the only remaining factor.”
*
Your disagreement with the specific question clearly means that you believe “Cos(A)” can be less than or equal to “Sin(A) x Cos(A)” from 0o to 90o… this is clearly wrong (this is the proper use of the word “wrong”) and can be verified by any one who can perform high-school math.
-

Your explanation that the Cos(A) factor will cancel-out is basically correct but the proof has large errors.
(One error is that “Sin(A)” is what remains… NOT “Cos(A)”, as you erroneously wrote in your earlier posting of 16/08/2008, and later corrected.)
The much larger error is one that requires thinking beyond the rudimentary of math.
What I mean is that we are making a comparison; NOT PRE-ASSUMING that there is equality, between the precession-torque and centrifugal-torque.
However the pre-assumption of equality may provide a tool to prove or disprove your theory…
-

Here is the PROOF of EROR in your calculation using your choice of notation.

Given:
(Fz x lz) = (m x R x Cos(A) x Wp**2) x (R x Sin(A))
(Fp x lp) = (Fi x li) = Fi x R x Cos(A)
When you assume equality between torques of precession and centrifuge (represented above), the result is a mathematical error (because the assumption is wrong as I intend to prove mathematically).

You propose that (Fz x lz) can equal (Fp x lp) as follows:
(m x R x Cos(A) x Wp**2) x (R x Sin(A)) = Fi x R x Cos(A)

Combining terms we get:
(m x Wp**2) x R x Sin(A)) = (Fi)

Solving for Sin(a) we get the following equation IMPORTANT to proving whether static-balance is possible:
Sin(A) = (Fi / (m x Wp**2) x R)

This means that if an equality between the two torques were possible then Sin(A) would be equal to the input force (Fi), divided by the magnitude of the centrifugal force (Fz) when the arms are fully extended!

Because we know that (m x Wp**2) x R is equal to the maximum centrifugal force (which occurs when the arm is fully extended), therefore (m x Wp**2) x R = max(Fz)

By substituting max(Fz) into the equation Sin(A) = (Fi / (m x Wp**2) x R) we get:
Sin(A) = (Fi / max(Fz))
…
We know that “Sin(A) CANNOT be larger than “1” (it is a mathematical IMPOSSIBILITY).
Therefore, if your equality-equation is true, it requires that “max(Fz) always has to be greater than (Fi); and we know this is NOT possibly TRUE.

This is my proof by “Reductio ad absurdum” using known equations placed into an equality that is a necessary component of your theory (as your theory is that (Fz x lz) can be equal to (Fp x lp)).

Please let me know me if there are any kind of ERRORS in my proof.
Perhaps you will show a proof of your theory.

By the way there is a reason why the centrifuge in a fast rotating-hub can appear to overcome the force of “precession” which results when the gyros begin a very slow spin. The reason is however based on a different set of equations involved during that stage… do you know what they are?

Best Regards,
Luis

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Answer: Luis Gonzalez - 24/08/2008 14:00:48
 Dear Harry,

I know today, that my hasty reduction ad absurdum proof is not correct.
However the rest of the posting does hold true.

Best Regards,
Luis

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Answer: Harry K. - 25/08/2008 12:47:05
 Dear Luis,

LG wrote:
"I know today, that my hasty reduction ad absurdum proof is not correct"

Yes I know! ;-) Though you have the opportunity for corrections before I write my response.

Regards,
Harry

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Answer: Harry K. - 26/08/2008 09:59:10
 Hello Luis,

Here is my answer to your posting from 23/08/2008 19:45:18.

LG wrote: "I see that on your subsequent posting of 17/08/2008 00:51:56 you corrected the glaring typing error in your previous posting (though the correction was not mentioned)."

I guess you mean my "glaring" mistake with cos(a) instead of sin(a) in my posting from 16/08/2008 15:11:34? - Oh boy, you have any bigger problems! It's obvious that this mistake was caused by hasty writing!

LG wrote: "This entire thread is about your theory but you have failed to prove it and have only presented snippets that would require others to embrace your theory by faith only. "

My dear Luis, I have explained in my very first postings in this thread all necessary wisdom about my theory of a balance point. In addition I have provided links to drawings on my private webspace for better understanding what I have stated about my theory here in this thread.
This provided information about my theory should be more than enough for "standard" educated people to understand this theory and to be able to find the behind mathematics to prove this theory.
Again, I have provided all necessary information in my first 4 postings in this thread, all other 96 postings were necessary to explain logical thinking and basic mathematics!

LG wrote: "I have said that I delve in matters of spin as a hobby (just for fun), and you are a claimed professional in the field (rare in this forum)."

I never claimed that I'm a professional in this field, but I only stated that I'm a mechanical engineer and therefore know pretty well how to calculate forces and torques acting in static equilibrium.

LG wrote: "Yet I introduce equations and factual concepts that are fresh to this forum (and to this thread)."

Your introduced equations were either well known or wrong.
Your "factual concepts" were either resulting from my concepts or were fresh and wrong.

LG wrote: "When I introduce logical insights (that cannot be disputed) you dismiss them as known facts, even though I have never seen them mentioned in this forum (I guess anything deduced logically from facts may be presumed to be already known). "

As stated before, your "introduced logical insights" were either well known (may be not here in this forum but science-based!) or logical wrong! Please name only one of your "introduced logical insights" which is new AND correct!

LG wrote: "It seems unusual to only addressed equations and concepts only in response to what I (and others) have stated, when you should be introducing concepts connected by equations, in support of your innovative theory on static balance (instead of only responding)? "

You began with stating faulty or confusing equations and thus I stated the corrected equations. Mathematic is common language which never lies and therefore a very good instrument for explanations, however, only if mathematics is understood from both parts!

LG wrote: "The word “WRONG” rolls-off with great ease in your recent responses, however the equations, in your last posting of 17/08/2008 00:51:56, confirm that my statements are correct.
(This has happened before in previous disagreements and we called them misunderstandings.)"

Nope, some of your equations were WRONG because you have missed to cancel cos(a) and thus your statements are WRONG because you did not understand the logical background behind this theory.

LG wrote: "The language-barrier may be hindering both the ability to understanding and the ability to express ideas clearly, and this appears to include the correct use of the word “WRONG”. "

Whenever you do not understand something, you mention "language-barriers". You problems are not caused by language-barriers but by the inability of consequent logical thinking.
When I write WRONG, I mean WRONG, FALSA, FALSCH! - Or should I better write UNTRUE, INCORRECT, FALSE, FAULTILY?

LG wrote: "I have provide significant indication that precession-torque is always greater than the centrifugal-torque, at all angles between 0o and 90o. Your posted equations have confirmed it as much as mine did (though in simpler notation), and the facts are recorded in this thread for anyone willing to do the high-school math (and able to discern through the postings). "

Don't make a fool of yourself! As stated before your "provided significant indication that precession-torque is always greater than the centrifugal-torque" is caused by WRONG understanding of my theory and basic logical facts!

LG wrote: "You confirmed my equations and thus proved me correct (though you claimed I was wrong), but have not yet proven your theory. I am sorry to have to say this but it is true. "

AGAIN, I did not confirmed all of your equations, because some of them are WRONG!


The rest of your posting regarding "reduction ad absurdum proof" is also WRONG and you have admitted this in your last posting as well.

In the moment I do not know, if it would make any further sense to continue with this thread. I could present the mathematical proof of my theory at once, but I really ask myself why I should do this? - That you later may claim "Yet I introduce equations and factual concepts that are fresh to this forum (and to this thread)." or something similar? NO THANKS!

I'm very disappointed about the progress in this thread and if nobody aside from you will ask me to continue, I will now call it a day...

Bye,
Harry K.





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Answer: Luis Gonzalez - 27/08/2008 00:44:49
 Hi Harry,

I am sorry you feel the way you do about our conversations.
Errors will always occur; different cultural patterns and attitudes will often cause friction.
Ill feelings often result from harsh words where insult may not have been intended.

Unless you intend to completely stop posting in this forum, we will continue to encounter each other.
We can stop putting insulting labels on each other, or we can avoid personality issues by writing in separate threads; thus removing the need to address the individual, and address only concepts, equations, etc in a less personal manner.

I will withdraw from this thread (start my own thread) and thus stop boring you with my way of thinking and my objections to your use of language.

I do prefer to keep my threads clear of other postings, as I can correct my own errors and evolve my own equations over time (with less strife), in my quest for gyro-propulsion.
Separate threads do not prevent others from writing counter thoughts to what I have written, but are less personal and will still allow everyone the freedom to speculate without personal repercussions.

Some may say this is a quest for fools, but I have never found true fools relentlessly seeking truth (I hope we said something worthwhile).

Farewell Harry K. and Best Regards,
Luis

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Answer: Glenn Hawkins - 05/09/2008 07:27:09
 EPITAPH

This whole thing has been both sad on one side and hatful on the other.
Except for Kidd’s appearance Not an ounce of charm or reason to exist is in any of it.
When science doesn’t abide in human kindness and purpose it’s use is dangerous.
I am very sorry for everyone involved.

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Answer: Luis Gonzalez - 20/09/2008 17:42:52
 Dear Harry,

If the (Fi) of a rotating hub is equal to the (Fz) of the hub, then it looks like I am correct about the static balance-point not being possible.

The static balance-point requires that (Fz) must be larger than (Fi), to compensate for the Sin(A) adjustment, so that torque (Fz x lz) can equal torque (Fp x lp).

Sorry Harry but this means that the “reduction-ad-absurdum” proof I presented on my 23/08/2008 19:45:18 posting in this thread, is correct and holds true.
You must have known this, or perhaps you think the (Fi) is not equal to the (Fz) in a hub with constant rotation?

Perhaps silence is best, as you don’t want to make a fool of yourself, especially in your profession.
On the other hand, I can afford to look a bit foolish occasionally during my amateurish quest in search of how gyro-propulsion may work (my professional credibility is not affected by how correct (or not) I am about spin physics).

Best Regards,
Luis

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Answer: Harry K. - 21/09/2008 00:19:00
 Hello Luis!

I see you're still struggling about satisfying answers? It seems we both have been the only one who were interested in this matter of gyro behavior and now it must be boring for you because you're are now alone, isn't it? :-)
Although I had decided to quit posting here if nobody aside you would be interested to continue in this thread, I nevertheless will give a reply to your last posting.

Luis wrote:
"If the (Fi) of a rotating hub is equal to the (Fz) of the hub, then it looks like I am correct about the static balance-point not being possible. "

I have never claimed that FI must be equal to Fz to achieve a balance point!? All I have stated is the necessity of equal TORQUES, i.e. Fp x lp (=Fi x li) must be equal to Fz x lz to achieve a balance point.
This is a typical basic misunderstanding from your side, which prevents progress in understanding of this theory!

Luis wrote:
"The static balance-point requires that (Fz) must be larger than (Fi), to compensate for the Sin(A) adjustment, so that torque (Fz x lz) can equal torque (Fp x lp).

Sorry Harry but this means that the “reduction-ad-absurdum” proof I presented on my 23/08/2008 19:45:18 posting in this thread, is correct and holds true.
You must have known this, or perhaps you think the (Fi) is not equal to the (Fz) in a hub with constant rotation? "

This is another basic misunderstanding from your side. Fi x li (TORQUE!) is derived by precession torque Fp x lp. I hope you can agree? Thus precession torque Fp x lp as well as Fi x li are affected by cos(a) and Fz x lz is affected by sin (a)!
For instance at an angle a=45 deg. sin- and cos-factor are equal. At angles under 45 deg. sin-factors are smaller than cos-factors and at angles above 45 deg, sin-factors are greater than cos-factors. Therefore your “reduction-ad-absurdum” proof is unfortunately pure nonsense. And once again, you have to consider always TORQUES and not only FORCES!

Luis wrote:
"Perhaps silence is best, as you don’t want to make a fool of yourself, especially in your profession. "

If i were you I would heed your advice! Some of your statements are not really glory...

Luis wrote:
"On the other hand, I can afford to look a bit foolish occasionally during my amateurish quest in search of how gyro-propulsion may work (my professional credibility is not affected by how correct (or not) I am about spin physics). "

Thus in your opinion you can state any nonsense here in this forum? Well, do what you want if you feel satisfied afterwards. :-)

As stated in previous postings here I have all the simple mathematics to proof my balance theory but I’m afraid you would not understand it anyway.

Have a good time and enjoy life, Luis!

Regards,
Harry K.



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Answer: Luis Gonzalez - 01/11/2008 12:35:40
 Hi Harry,

In response to your posting of 21/09/2008 00:19:00 in this thread:
Finding the answers is the fun part; explaining them is the frustrating struggle. I can’t get even you to understand what I say (even though we agree on some basics); but it’s common in this subject.
I assure you that plenty of people are interested in reading about matters of gyro behavior, but most are unsure what to say, or don’t understand.

Harry wrote:
“I nevertheless will give a reply to your last posting.”
*
What makes you think this is my last posting… I am just getting started.

Harry wrote:
“I have never claimed that FI must be equal to Fz to achieve a balance point… This is a typical basic misunderstanding…”
*
Yes Harry, it’s your misunderstanding…
You are confusing my proposed conjecture, with what you think I am saying about you.
In the statement:
"If the (Fi) of a rotating hub is equal to the (Fz) of the hub”
I am simply proposing a conjecture (do you get it?).
Then I conclude that, if the proposed conjecture is true, then the static balance-point is not possible!
Just a bit of not-too-subtle English, that’s all.
(The conjecture is not really true, but it provides a test to agree or disagree about.)

Harry wrote:
“Thus precession torque Fp x lp as well as Fi x li are affected by cos(a) and Fz x lz is affected by sin(a) … For instance at an angle a=45 deg. sin- and cos-factor are equal…” etc.
*
Sorry Harry but once again you forgot to say that Centrifugal-torque (Fz x lz) is affected by BOTH Sin(A) and Cos(A)… (This is a relevant statement even if Cos(A) can be cancelled when the comparison is performed.)
It is reassuring to see that even pros make inaccurate statements, about important equations.
While it is correct to say that Cos(A) affects precession torque (Fp x lp) as well as (Fi x li)…
However your statement hides the fact that the overall centrifugal-torque (Fz x lz) is affected not just by “Sin(A)” but rather by “Cos(A) x Sin(A)”…
We have had this argument before in this thread on 11/08/2008 18:54:11 through 23/08/2008 19:45:18, and then you kept changing the focus of the conversation…
In that posting I said:
(1) - (Fp x lp) is affected by a factor of “Cos(A)”, but…
(2) - (Fz x lz) is affected by a factor of “Sin(A) multiplied by Cos(A)”

Harry responded:
- Fz is affected by factor “Cos(A)”
- lz is affected by factor “Sin(A)”

And you claimed our statements to be different, and that mine was wrong, etc!?
It doesn’t matter how we parse it Harry… Fz and lz are factors of centrifugal-torque, and when we unfold the whole equation we find that the overall centrifugal-torque is affected by a factor of “Sin(A) x Cos(A)”.
And the precession-torque is affected by a factor of just “Cos(A)”.
And yes, after combining like terms, to cancel Cos(A), the side that started out being the centrifugal-torque (Fz x lz) is still affected by a factor of Sin(A)!!.
Many of us know that when these 2 torques are placed in a relational equality, the Cos(A) factors cancel out, as do other factors. But this does not negate the validity of the previous statements… I think it was just your way to change the conversation and to cast doubt regarding the core of this argument, which is:
CAN (Fx x lz) equal (Fp x lp) while spin is faster than hub rotation?? (I.e. is a viable static balance-point possible??).
We know the answer is NO!

The item of real importance is that your static-balance-point is an erroneous conjecture, which you yourself have expressed severe doubts about.
****
The simplest clarification is that “Precession-Torque” is equal to the product of mass times hub-velocity times spin-velocity of the gyro…(Fi x li) = (M x Vi x Vs)…
While “Centrifugal-Torque” is roughly equal to mass times the square of the hub-velocity with an adjustment of Sin(A)/Cos(A)…so that (Fz x lz) is approximately equal to M x (Vi**2) x Sin(A) / Cos(A).
(I hope this last set of statements does not confuse too many readers. If you need to reconcile all the equations simply expand the equation “(Vi**2)…etc, and the trig terms will work themselves out.)
***

Anyway we look at these equations (coherently) tells us that all evidence indicates that your static balance-point is only possible under conditions different from what you claim, a figment caused by isolating some factors, and failing to look at the overall relevant picture.

Other facts we can look at are:
The net “Sin(A) x Cos(A)” factor of Centrifugal-Torque is always a fraction (less than “1”) and is always smaller than the “Cos(A)” factor of Precession-Torque.
This is TRUE when we look at the torques separately; and the BASIC TRUTH remains even after we cancel out Cos(A) and any other like-factors in a comparing equation.
Comparing (Fp x lp) and (Fz x lz) through an assumed equality lets us cancel out like-factors and end up with an equality that depends on Sin(A) as the only factor dependent on the angle of elevation.
From this perspective, the defining question is whether at any angle the equality equation can be satisfied??
The bottom line is that at all angles including above, below, or at 45o, Sin(A) is less than one, which is also true of Cos(A) and about the product of Sin(A) times Cos(A), which is even smaller then either Sin(A) or Cos(A).

Harry wrote:
“Some of your statements are not really glory...”
*
We both have some excellent and some not so glorious statements that have made both look a bit foolish on record in this forum (I guess it was your turn this time again).

Harry wrote:
“…in your opinion you can state any nonsense here in this forum?”
*
Unlike some, I can introduce conjectures that are not mentioned in most Physics text-books (and are new to this forum), because I am not afraid of letting out any great secrets, or of being judged by my professional colleagues; however be assured that I always intend on winning my arguments through valid logical statements.

Harry wrote:
“I have all the simple mathematics to proof my balance theory but I’m afraid you would not understand it anyway.”
*
Based on your own statements in previous postings, you don’t really fear that I would not understand… quite the contrary; we know what your fear is, that I (and others) DO understand, and that we may see through the errors in your basic beliefs. Your bluster is not believable, and your errors are indelibly recorded in this forum.
I am sure you have all the necessary Physics books to make you feel confident, but I’m also certain you have not proven your static balance-point theory because any such proof is in error. This leads me to think that either you seek to mislead on purpose or you are unable to understand your books fully in this subject (you wouldn’t be the first to be wrong).
I do however agree that whatever proof you have, could not be understood by anyone including yourself. When you finally discover your error, you will see what I am talking about.
-
I think our continuing disagreement is largely caused by our egos. In your posting of 02/06/2008 18:44:19 you said:
Harry Wrote:
“You know, I also have big doubts about the truth of this theory. I beliefe I'm "blind" to find possible errors by my own and therefor I would be happy if somebody could reveal the errors.”
*
I have tried to reveal your errors, and have certainly made mistakes along the way (it is a tough nut to crack), we have managed to annoy each other but the facts have a tendency to emerge with a little persistence.
Unfortunately we are no longer viewing each other in favorable light. I have tried to take the conciliatory road before but it has only earned me condescending brusque responses.
You can keep your books though I know they contain much valuable information. I will continue to struggle on my own…posting items that may not be fully correct and then fixing my errors until I get it right. In my favor is that I enjoy the subject and I am sufficiently persistent. I wish I had more time to spend on it because it is a fascinating subject.

I hope you find at least half as much enjoyment as I derive from the struggle in this quest.

Thank you for the good wishes; I wish you the best and hope you succeed in your quest to derive power from spinning objects. Frankly I just don’t believe that it’s possible to get more energy than is put into a system!!!.

Regards,
Luis G

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Answer: Luis Gonzalez - 01/11/2008 12:44:00
 Hi Harry,

Regarding your posting of 20/10/2008 21:05:05 in thread http://www.gyroscopes.org/forum/questions.asp?id=967:
Your equation, which states that “(Ti) is equal to (Vp/rp x Vs x M x r/2)”, is in error because it is adding unnecessary parameters to a well established equation for precession.

The known, well accepted equation states that the velocity of precession is equal to the input-torque divided by spin momentum.
In other words: Vp = (Ti) / (Vs x M) is a correct equation.

Your equation claims that “Ti = (Vp/rp x Vs x M x r/2)”, and is in effect a modification to the above well known equation; let me show how that is so!.
When we rearrange the terms in your equation to solve for velocity of precession (“Vp”), we end-up with:
Vp = 2(Ti x rp) / (Vs x M x r)
That just DOES NOT sit well against the well established equation, which is Vp = (Ti) / (Vs x M)!
The parameters (2 x rp / r) are extraneous and not necessary.
Therefore, as you would unceremoniously say, your claim appears “wrong”!
Otherwise, how do you justify the extra factors?

If this extra factor were true, that would mean that precession’s velocity would be larger by approximately twice the ratio of the “hub-radius” over the “gyro-radius”, which can turn out to be from 6 to 10 times larger then the real velocity of precession!
I know this difference in calculating precession’s velocity would have been noticed in many tests and experiments during the last 100+ years.

Sorry Harry but if we allow the cavalier aggregation of unexplained factors we can make the equations support any conjecture no matter how outlandish!
--
IMPORTANT!
I also want to say that your claim about the maximum angle of balance (as 45o) is based on erroneous calculations.
It looks like the value of your calculations keeps shifting, thus proving that there is a serious error in your analysis; that puts you in ground that is not much different from other contributors to this forum.
Though you prefer to think of yourself as a better prepared in regard to gyro information, your admitted reversals in opinion point at your errors and indicate that you are just another average contributor to this forum, and not any better prepared in gyro matters than me or many others here.

Here is a prime example.
~Harry wrote:
“My new corrected equation states also, that the maximum angle of balance cannot exceed 45° above horizontal plane.”
*
***I believe the above statement is just as wrong as when you claimed that the static balance-point was at about 80o.
I know why you made the statements; they are the result of your errors in calculation.
It is true that the angle-driven parameters of the centrifugal-torque start decreasing once the angle increases above 45o (after having increased from zero at 0o, and maxed-out to 0.5 at 45o).
(Sin(45o) x Cos(45o) = 0.5.)
***
However, the centrifugal-torque’s angle-driven parameters CONTINUE to CONVERGE with the angle-driven parameter of the precession-torque!!
(Do you get it?)
(A graph of Cos(A) and of Sin(A) x Cos(A) should clarify the meaning of my last statement.)
It means that when a balance angle is created (through appropriate means) it can indeed exceed 45o (sorry).
This is something I tried to explain to you on my posting of 11/08/2008 18:54:11, in this thread, but you responded that I was wrong (to which I took exception and things went downhill after that).
***

~Harry wrote:
“I have studied your equations … Although I cannot follow with some of your derivations,…”
*
All my equations in that thread (thread id=967) were simple enough for any engineer to follow; perhaps I made a typo…which steps were you unable to follow?

I don’t expect you will answer directly any of my pointed questions, and that is okay. Though our disagreements can at times become unfriendly, I still value some of your out-of-the-book technical opinions.

Regards,
Luis G

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Answer: Harry K. - 03/11/2008 14:30:55
 Hello Luis,

Strong words from a very self-assured guy, WOW! - However, you should be careful in what you are writing here.
I had already admitted in your "private-read-only-thread" that I have made a mistake in my calculations of the balance point. But this does not mean that YOUR calculations are correct, never ever...

I give you an example:
You wrote:"In other words: Vp = (Ti) / (Vs x M) is a correct equation"

No it is NOT a correct equation. Here is the derivation from the CORRECT equation:
Wp = Ti / LG

Wp - angular velocity of precession
Ti - input (tilting) torque
LG - angular momentum of spinning gyro

LG = JG * WG

JG - mass inertia of gyro

Mass inertia for disk shape: JG = (mG * rG * rG) / 2

mG - mass of gyro (disk)
rG - diameter of disk mass

Wp = Ti /( JG * WG)
Wp = Ti / (((mG * rG * rG) / 2) * WG)
Wp = Ti / mG / rG / rG * 2 / WG

Wp = vp / rG

vp - circumferential velocity at rG of precession
vG . circumferential velocity at rG of spinning gyro

(vp / rp) = Ti / mG / rG / rG * 2 / (vG / rG)

-> Ti = (vp / rp) * mG * rG * rG / 2 * ( vG / rG)
Ti = vp / rp * mG * rG / 2 * vG
-> Ti = vp / rp * vG * mG * rG /2

QED
That's maths, any logical doubts?

Your (wrong) equation disregards the geometric shape of the gyros as well as the radius of spinning mass!
Also I have advised you several times to use angular velocity (W) and not circumferential velocity (v). The equations will become unnecessarily complex by using circumferential velocities instead of angular velocities!

I would advise you to read in this thread to learn more about basic equations for gyro calcultions. In the moment I have less time but I will continue in this thread as soon as possible.

http://www.gyroscopes.org/forum/questions.asp?id=981

Regards,
Harry K.

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Answer: Harry K. - 16/11/2008 17:35:20
 Come on Luis! I'm still waiting for your answer. I have proved you wrong by using basic maths equations. Although I know that logical way of thinking is not your passion, I would be curious to read your simple truths and logical reasoning regarding this issue.

I'm waiting...

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Answer: Harry K. - 20/11/2008 10:49:11
 Dear forum,

BYE

Harry

P.S. Luis, I'm no longer waiting for you answer.;-)
Your flutter theory is crap. Don't waste your time!

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