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7 May 2024 02:52
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Question |
| Asked by: |
Luis Gonzalez |
| Subject: |
Gyro Knowledge – Does the rate of precession change? |
| Question: |
Given:
A gyroscope spinning at constant angular velocity is attached from the opposite end of its axle to a vertical drive shaft via a pivot free to swing up and down (Type “A” Laithewaite system, single or double).
Precession wp = 0.2 cycles per second under 1G (32 feet per second per second).
Move the system to a gravity free environment.
Next, rotate the vertical shaft at 0.2 cycles per second.
Assume gyro is driven while rest of device remains stationary (by use of in sync counter systems etc.)
Questions:
At what rate will resulting precession swing the gyro upward?
Will rate of precession be uniform through entire arc, until it reaches a balance position at the top?
If the rate varies, will it be faster when the gyro axle is near horizontal, or when it is near to the vertical balance position?
Thank you,
Luis |
| Date: |
26 December 2005
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Answers (Ordered by Date)
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| Answer: |
Glenn Hawkins - 28/12/2005 16:59:00
| | | Hello Louis,
Maybe this will help a little. Consider that you first force your gyro to precess in gravity at a constant horizontal plane. Any more added force than that will cause the gyro to lift. The difference in a gravity free environment would seem to be that because there is no energy necessary to overcome gravity, all the energy used to force precession should be converted to lift. Any rate should lift about, 1:1.
I think the rate of precession should be uniform so long as the energy source of input for lift remains uniform. Looking down on your apparatus you should notice that the circumference of beginning precession (45o tilt) is small. As the gyro rises to (90o tilt) you would notice that the circumference of precession had grown larger. As the rise continues toward the top (135o tilt) the circumference of precession will have constantly decreased. If one were to set a melon on the table and slice it into horizontal pieces each slice would, from bottom to center would grow larger and each slice from center to top would progressively decrease in circumferences. The slices would be good to represent a visual of changing precession circumferences.
Back to the gyro. Note that if as I think precession orbits remain the same as in (RPMs), the distance traveled in each preceding lift circumference would be shorter from bottom, longer at center, and back to shorter at top, even though the RPMs remained the same.
Here is the clencher. If you perceive there is force to be had from precession then the speed of mass and distance traveled are relative to how much force will be available. The reason the gyro is able to precess at the same RPMs though the distance traveled is different is a matter of alternating leveraged forces.
One may perceive the leverage changes this way. If he draws a pedestal and then a gyro at three positions while keeping the axel the same distance (stick drawings will do, just a few marks) he may better see leverage differences. When the stick drawn gyro is assumed to set on a table (7: clock) it’s ‘horizontal’ distance from the center of the pedestal may be 2”. When the gyro is drawn and aliened to (9: o’clock) the ‘horizontal’ distance of gyro to pedestal may measure 3”. At (11: o’clock) the ‘horizontal’ distance should be back to 2”. Perhaps in this way we may better realize that lesser to greater to lesser leverage is how the gyro compensates for different mass speeds at different elevations though the RPMs should remain the same. Differing amounts of force are applied.
Conclusion: Precession rate should not change. In zero gravity all torque force should be converted into lift, or rather into an upward curving force.
Well, I’ve found a way to escape from my drawing board for a few minutes. Now if I can only force myself back to it.
Good Luck, Louis,
Glenn Hawkins
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| Answer: |
Luis Gonzalez - 29/12/2005 15:44:57
| | | Glenn,
Regarding the 1:1 ratio of conversion to lift; the angular velocity of the flywheel does affect the rate of precession. When the flywheel spins faster the rate of precession is slower. Therefore, given the same torque from the vertical shaft, a faster spinning gyro will have a slower rate upward precession then a slower spinning gyro. Obviously the response can NOT be 1:1 in all cases given the same torque (i.e. different rates of spin will produce different rates of precession).
The horizontal melon slices provide a good visual, but the resulting circumferences do not represent the direction of precession because the vertical shaft’s sideways torque causes precession UPWARD, not sideways (i.e. it is not the same as a gyro on a tower, a common source of confusion).
Your point about changing leverage (of torque, I assume) is valid and should affect the rate of precession accordingly. However, it contributes to a changing rate of precession and not at all to maintaining precession constant.
You may want to consider that at positions where the gyro axle is less horizontal, CENTRIPETAL force exerts a stronger influence on the rate of precession. When you combine the fluctuating centripetal force with the fluctuating force due to changing “leverage” of torque, it could produce a zone where the precession is uniform, but only on gyros that have a relatively smaller radius flywheel.
Systems using larger diameter flywheels introduce a new family of variables. First of all, the rim of the flywheel extend over more than just one latitude along the upward precession; the leading and trailing edges will be at different latitudes (your clock positions, 7;9;11 or slices of melon) at any given point in time. Second, the mass of the leading edge, in larger flywheels, will cross over the upper balance point affecting the magnitude of the centripetal force and possibly, to some degree, affecting the “leverage” effect. I seriously suspect that all variables involved will NOT change in perfect sync to maintain uniformity of precession, especially when using flywheels of larger diameter.
Speculation is limited by our perception and imagination until a mathematical model is used and verified by experiment (or visa versa). I suspect that Euler’s equations will provide what I am looking for, but I do not have the training to understand which, where, and how to use them if they exist (this type of help would be most welcome).
If equations that cover the subject at this level of detail do not yet exist, then it is a wonderful opportunity for bright young minds to discover them. Are there any such minds out there that will rise to the challenge?
Thank you for your ideas and imagery which were helpful to expanding my understanding of the scenario.
I am curious what type of drawing do you do on your board? Is it patent drafting?
Thank you, Luis
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| Answer: |
Glenn Hawkins - 29/12/2005 19:11:11
| | | Lous,
Your first paragraph is a given, but I thought everyone knew that. You did not ask about variables in disk diameter, variables in mass, variables in mass displacements, or variables in spin speed.
Louis, one must use exactly the same gyroscope and gyroscopic conditions to compare operations in gravity, to operations in zero gravity if he is to make a sensible comparison of differences.
In any case I said, “I think the rate of precession should be uniform so long as the energy source of input for lift remains uniform.” That is an all encompassing and clarifying statement! All the energy (in put) that goes into lift includes angular momentum, don’t you see.
Why would you wish to find an argument? You seemed sincere. You seem to work hard and you’re shearing. You ask two questions and I answered two questions taking the trouble to give you careful mechanical reasoning.
Tests on the functions of the propulsion system on my drawing board have been measured with spring scales to be, applied force to thrust, 1:1. This also seems also the right answer to your question, but you may note that I said, “I think” and “should be”. I think I’m through with this.
So long Louis,
Glenn Hawkins
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| Answer: |
Luis Gonzalez - 30/12/2005 05:16:41
| | | Glenn,
We must be able to agree to disagree on this complex subject limited by language, length of space to write succinctly, and different angles of perception.
It is important that we evaluate what is written and provide constructive criticism where our unique perception tells us is necessary.
I hope that one of us, or a team of us, in this forum make the big breakthrough needed to make our efforts legitimate in the eyes of the world including science and investors.
Thank you, Luis
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| Answer: |
chris jones - 02/01/2006 00:03:39
| | | Will rate of precession be uniform through entire arc, until it reaches a balance position at the top?
I think the top you refer to is only available with gravity
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| Answer: |
chris jones - 02/01/2006 00:03:43
| | | Will rate of precession be uniform through entire arc, until it reaches a balance position at the top?
I think the top you refer to is only available with gravity
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| Answer: |
Luis Gonzalez - 06/01/2006 03:35:14
| | | Chris,
The term “top” is just a reference to the portion of the vertical shaft that occupied to uppermost position before the system was moved to the gravity free environment. I chose to maintain the same reference to avoid long drawn explanations (Is there another name that you would prefer to call it?). Maintain the image of the device (while it was on the surface) and use the reference points in the gravity free environment.
In the gravity free environment, the acceleration is provided by the torque of the vertical shaft (this acceleration is not directed in the same “downward” direction resulting from gravity on the surface.) The new acceleration is round about sideways (created by the turning of the “vertical” shaft) which will cause “upward” precession, because it is applied in the same direction that the device would have turned in precession under gravity (if not familiar please review the Laithewaite Christmas lectures).
Thank you,
Luis
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