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23 November 2024 15:21
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Question |
| Asked by: |
Blaze |
| Subject: |
A Theory on Changing Magnitudes and Changing Forces during the Drop |
| Question: |
This is a theory on changing Magnitudes and Forces during the Drop when starting (or the Rise when stopping) an overhung gyro. The theory is for a gravity “powered” gyro system but would also apply to any system that has a mechanically generated “Down Force”. As always, input is appreciated.
Premise:
1) During the Drop the precession speed starts from zero and increases (not necessarily linearly) to steady state speed over some short but measureable period of time.
2) All reactions are instant but not equal in Magnitude until steady state precession speed is achieved.
3) The Magnitude of the Up Force (opposite of gravity) is directly proportional to the speed of precession, therefore, as precession speed increases in magnitude, the Up Force increases in magnitude.
4) While precession speed is increasing, the gyro wheel is being driven by a Force, however, that Force is being split with the Up Force. The sum of the Force driving precession and the Up Force is always equal to the Down Force or gravity.
5) When steady state precession speed is achieved it is a motion only with no Force. All the Force is in the upward direction, the Up Force.
Discussion:
Point 1):
I think that it is generally agreed upon that during the Drop precession speed increases from zero to steady state speed. Whether or not it is linear does not matter to the theory being proposed here.
Point 2) and point 3):
The reaction to gravity is precession motion. The reaction to precession motion is an upward force resisting gravity. These reactions are instantaneous however their magnitudes are not instantly equal to the magnitude of the gravity force. This is most easily seen when looking at the Up Force. An argument can be made that the precession speed determines the Magnitude of the Up Force. It is well known that if you increase the precession speed to faster than steady state speed the gyro wheel rises. If you reduce the precession speed to slower than steady state speed the gyro wheel falls. This upward or downward movement is a result of a changing Magnitude of the Up Force. Therefore, since the precession speed changes from zero at the start of the Drop to steady state speed at the end of the Drop, the Up Force must also change from nearly zero at the start of the drop to equal to the force of gravity at the end of the Drop (neglecting friction).
Point 4) and point 5):
Let’s first start with the last statement of point 4) which basically is saying that the resultant Forces are equal to the input Force (gravity or Down Force). When steady state precession is achieved the Magnitude of the Up Force must equal the force of gravity (neglecting friction) or the gyro wheel would fall. From the discussion of points 2) & 3), it would seem that the Magnitude of the Up Force varies with precession speed. So during the Drop, where is the rest of the resultant of the force of gravity when the Magnitude of the Up Force is not yet equal to the force of gravity? I believe it is the Force that is accelerating the gyro wheel to steady state precession speed. Since the sum of the resultant forces equals the input force, this would mean that at the very start of the Drop, the precession Force would be almost equal in Magnitude to the force of gravity and because the precession speed is very small at the start of the Drop, the Magnitude of the Up Force would be very small as well. The total Magnitude of the resultant of the force of gravity would be the sum of the precession Force and the Up Force. Near the end of the Drop time, when the precession speed is almost at steady state speed, the precession Force would be very small and the Up Force would almost be equal to the gravity Force. When steady state precession is achieved, the precession force would be zero and the Up Force would be equal to the force of gravity, which brings us to point 5).
Of course this would imply that there is a horizontal force on the pivot during the Drop time and that is correct but that horizontal force only exists during the Drop and NOT during steady state precession. This horizontal force is noticeable in a very small movement of the pivot only IF the pivot is allowed to move and only IF the spinning speed of the wheel is very low. If the wheel speed is spinning at a high rate of speed, the horizontal movement of the pivot would too small to be noticeable even if the pivot were allowed to move. For example, consider a system where the wheel diameter is 5 inches and the distance from the pivot is to the center of the wheel is 2 inches. For a wheel speed of 500 rpm this would give a Drop time of about 24 milliseconds (about 1/41 of a second) and a travel distance during the Drop of about 1/8 inch (noticeable), but with a wheel rpm of 10,000 the Drop time is only 1.2 milliseconds (about 1/818 of a second) and the travel distance during the Drop is about 1/3467 inches (definitely not noticeable).
Point 5):
From the discussion in point 4), when steady state precession is achieved, there is no longer any precession Force to accelerate the gyro wheel and therefore precession speed is constant like it is coasting (neglecting all frictions) and the Up Force is equal and opposite to the Down Force (gravity).
So, why is this important? Because, although it still doesn’t explain why there is continued steady state precession movement without a force to continue to make it move, it does give a glimpse of why mass movement happens and why momentum exchange exists when precession speed deviates from steady state. And after all, it is all about momentum.
Your thoughts and comments are appreciated.
Blaze
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| Date: |
20 May 2012
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Answers (Ordered by Date)
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| Answer: |
Luis Gonzalez - 20/05/2012 04:19:15
| | | Hi Blaze,
I am on board with most of it except for the “Up Force”.
I suppose the Up Force could be fitted in to take the place of the notion that the mass of the flywheel somehow Shifts to the Pivot Point. In any event, you should not have to need both.
For my money you should not need either UP or SPP, except as an engineering rule of thumb.
All that should be needed is to follow a mass particle on the flywheel, along with its opposite mass particle (opposite mass particle being one located 180 degrees on the flywheel).
However my approach is too convoluted to be explained clearly.
I have been wrong before, perhaps this is one such occasion.
Best Regards,
Luis G
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Ram Firestone - 20/05/2012 07:34:23
| | | I'm not sure I understand all your points ... however..... if Lathwaite’s theory is in any way correct I would tend to say it must happen only during the drop phase due to the conservation of energy problem I keep harping on. During the drop is the only time potential energy is being lost therefore that is the only opportunity for the gyroscope to provide energy for mass transfer. The only question is does all the lost energy go into precession or does some of it go into Lathwaite’s mass transfer?
Keep in mind there is also nutation which is basically a spring effect. Precession happens in all directions. So the first order precession can actually cause a second order upwards precession which causes a backwards precession and so forth. Nutation often causes the gyroscope to go in little loops as it continues on its major precession path. Even the earth has nutation.
In any case If we assume for a second mass transfer is correct (at least during the drop) then it would seem the goal is to extend the time the gyroscope is dropping. With a gravity driven gyroscope this seems difficult if not impossible however it should be doable by using magnetically driven precession. But again this is (a) only a wild theory at this point (b) breaks Newton’s third law and (c) would have to be CAREFULLY experimentally tested to see if it’s true or not. At least it is something that has a chance of working however small because it’s possible to reconcile conservation of energy with this theory.
This all gets back to Laithwaite’s eureka experiment. Did he really see what he claimed or was he mistaken and his results where the result of too much friction in the experiment. If he really did see mass transfer it’s possible that he never quite found the solution because he never considered it might only happen during the drop phase.
If you want to try experiments like this this keep in mind FICTION IS YOUR ENEMY as it can make you think you have found something when you haven’t .
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| Answer: |
Blaze - 20/05/2012 15:30:21
| | | Actually Luis, the "forces" view and the "particle on the flywheel" view are really just two different ways of looking at the same thing. The "forces" view is a more of a macroscopic way of looking at it and the "particle on the flywheel" is more of a microscopic way of looking at things.
All "forces" cause rotations in an overhung gyro. The Up Force is caused by a rotation around the vertical axis through the flywheel but the Up Force causes a rotation around the horizontal axis through the flywheel and since the flywheel is mechanically connected to the pivot through the shaft it causes the downward force on the pivot and thus weight transfer to the pivot (well that is the cursory explanation anyway).
I agree with you that "All that should be needed is to follow a mass particle on the flywheel, along with its opposite mass particle (opposite mass particle being one located 180 degrees on the flywheel)." If you look at it from this microscopic view you can "see" everything.
Blaze.
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| Answer: |
Glenn Hawkins - 20/05/2012 22:55:37
| | | Hello Blaze,
Before I began:
This condition was once important to my designs. I worried that the pivot would move in the opposite direction during acceleration into precession. Then recently I created actual inertial acceleration that coasted about two feet. I did it three times and was satisfied. That is why I forgot I had learned that the pivot jerks sideways sometime, under certain lose conditions. Harry K' suggested Alzheimer. Engineers can be so cruel. : ) Blaze argued with me till I finally realized he was right and that I had once been right. Propulsion created or not, the pivot will jerk oppositely under acceleration.
Now to begin:
The speed of gravitational waves in the general theory of relativity is equal to the speed of light in vacuum. Now we see a hint why Momentous was for all practical purposes correct, though I argued against him and against you too, Blaze. Momentous said the drop into precession is about as instantaneous as you can get. That is how fast gravity influences a body. However the actual speed of the body into gravity is only 32.2 feet per second. Now postulate the the acceleration from -0- to 32.2 ft. per sec. and the acceleration fall of 1mm. That would be 1/805 of sec. from -0- to full state.
We could probably do away with this discussion now, because gravity is extremely weak compared to the greater electromagnetic force generated by a motor. By this we may know that powered precession can be thousand of times quicker than the near installations drop into gravity. In our pursuit we haven't got a need to work in the exacting, infinite numbers possible in mathematics.
The problem I mention never actually existed, but in my mind. I worried that an electric motor would exert continuously greater tilting force (which it can). I worried that the pivot would be subject to continuous opposite rotation increasing exponentially, therefore increasing the movement and force of the pivot to go oppositely.
However it is clear now the jerk is such a tiny distance compared to precession distance, it can for my part be eliminated from consideration, but there is more reason to ignore it. Consider that the tilting force is graduated ( 100 ) times from zero to steady state. Each time precession is swung faster, the acceleration is paid for and virtually little else is owed to keep it precession at that speed. It is the same for each graduated increase. The exerting force does not have to be more than ( 1 ) through the entire graduated force and speed may increase (100 ) times.
Back to this post heading:
I am for realizing the drop in a great gyroscope is real, but so near instantaneous that a distinction is without practical use and consideration toward our designing and building.
Glenn,
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| Answer: |
Blaze - 21/05/2012 06:59:54
| | | Hi Glenn. Thanks for your input. By great gyroscope I think you mean large, or really big. So consider the following system.
Pivot to wheel distance of 10 feet
wheel diameter of 5 feet
wheel speed of 500 rpm
wheel design is solid of uniform thickness throughout
This system gives a drop time of 0.61 seconds and a travel distance during the drop (the jerk distance) of 6 feet.
So, not all that instantaneous and certainly a significant distance.
Now make the wheel about 20 tons and park the whole thing on a spacecraft and you have a size that could be used for propulsion. Yes I know this would still be inchworm movement but that is not the point. I guess it really depends on what you want to do with that kind of movement.
Blazing on,
Blaze
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| Answer: |
Glenn Hawkins - 21/05/2012 12:29:50
| | | Hi Blaze,
How's your morning going?
None of that is realistic to me and I stand my ground.
Well ok. Look, you can increase the RPM and reduce the size of the wheel and build just as much angular momentum.
You can reduce the shaft length and speed the four stroke action, which is what it will be.
The greater distance traveled in precession in your example is a total waste of energy if you've been following me.
The shorter shaft means, quicker action and quicker repetitions of actions.
I believe I gave you good and carefully laid out information.
Take it easy,
Glenn
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| Answer: |
Blaze - 21/05/2012 18:14:17
| | | By the way Glenn, I like your idea of applying 1 step at a time for 100 steps. Very interesting.
Blaze
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Blaze - 21/05/2012 18:47:47
| | | "although it still doesn’t explain why there is continued steady state precession movement without a force to continue to make it move"
Actually, I take that back. This does explain why there is continued movement without a force to continue to make it move. Consider that once steady state precession is reached the weight of the gyro wheel is "just sitting on top of the pivot" and that, neglecting frictions, the gyro wheel is "coasting". You put these tow together and you have the answer (you would have to look at the mass particle microscopic view to see the complete answer but the wheel sitting on top of the pivot and the wheel coasting is basically it). There is no reason for it to stop except for the tiny amount of friction from the pivot, and windage.
By the way Glenn, you should like the concept of "coasting" as I got it from you in a much older posting.
Blaze
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| Answer: |
Glenn Hawkins - 21/05/2012 22:40:10
| | | Hello Blaze,
Thank you. You are a cleaver guy.
I made a misstatement: Please forget it,
"The exerting force does not have to be more than ( 1 ) through the entire graduated force and speed may increase (100 ) times."
Instead have this if you will.
In a graduated scale, the tilting force of 1 unit of power may be repeated 100 times in rapid secession, but the speed of precession will remain equal to only 1 unit of power.
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Glenn Hawkins - 21/05/2012 22:41:52
| | | Harry and I should not be surprised at my misstatements. Harry thinks either I, or he is losing our mind. Ask him. No wait. If it is he, that is affected asking him wouldn't do any good. (Hi Harry) Perhaps neither of is certain. He points out my inability to recognize names, while at the same time he has taken to singing off this way, “Mr. Alzheimer. :-)))”. See what I mean? I am worried about us. Without joking though, dementia is a very frighting thing. I have seen it. I try to be respectful to the people who have it.
Glenn,
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| Answer: |
Blaze - 21/05/2012 23:08:18
| | | "Instead have this if you will.
In a graduated scale, the tilting force of 1 unit of power may be repeated 100 times in rapid secession, but the speed of precession will remain equal to only 1 unit of power."
That is the way I interpreted it Glenn.
Blaze
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Blaze - 21/05/2012 23:26:09
| | | Actually, thinking about it more Glenn, I wonder if we are not working on basically the same idea, only I am looking at it in a "reverse" manner to the way you are. If this thing can in fact be done, there is very likely more than one way to accomplish it.
Keep blazing on,
Blaze
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