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The Gyroscope Forum |
6 May 2024 11:43
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Welcome to the gyroscope forum. If you have a question about gyroscopes in general,
want to know how they work, or what they can be used for then you can leave your question here for others to answer.
You may also be able to help others by answering some of the questions on the site.
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Question |
| Asked by: |
Luis Gonzalez |
| Subject: |
Challenges and solutions in accelerated systems |
| Question: |
Sandy has made it clear that his design does not attempt to make use of “thrusting gyros downward” subsequent to significant precession swings. His design, Sandy states, will make use of “accelerated mass” (I believe Sandy is referring to a mass that has acquired a large momentum). He further states that by the time precession has kicked in (about and after the saturation zone), the “accelerated mass” has dissipated. Therefore his gyros will need to be spinning very slowly (most likely intermittently OFF/ON/OFF/etc, quickly but in sync) while the system is rotating as fast as possible rate to produce maximum “accelerated mass” i.e. momentum.
A most ingenious part of this design is the simplicity of the main concept. The mass/flywheels (with a very high momentum) should be kicked in a direction 90 degrees from the very fast rotation of the system (caused by on/off action), and then stopped abruptly by a stopping-component (that should be mounted on springs or such to prevent damage). This component should also not interfere with the speed of rotation of the system; it should only stop the motion, which was created by quasi-precession (at 90 degrees) and which should also have a very large momentum. (If it was true precession then the quantity of momentum in the direction of precession would be questionable or week at best).
The short story is that the kicked-up, high-momentum-mass encounters a stop and transfers its momentum to the stopping-component causing it to accelerate. The beauty of the design is that the kicked-up, high-momentum-mass is not expected to have a reaction at 180 degrees (except when it encounters the stopping-component) because the kick-up is itself the reaction at 90 degrees from the system’s rotation as it interacts with the flywheel’s intermittent spin.
Will it work?
In theory it should if in fact there is no reaction from the kick-up at those low spin rates of the flywheel (experiments show positive results).
So where is the snag?
With high spin flywheels, it’s Nitro’s first law that causes havoc in most designs.
With intermittent and/or slow turning flywheels the challenge is in transferring the “accelerated mass” i.e. high momentum to the kick-up.
Other, not as imminent, challenges include how to stop the kick-up and bring the mass back for the next kick without undoing the initial effort at least in part (degree of vibration).
Where is the solution?
Mounting gyros in a radial configuration, from the center of rotation of the system, sets a scenario where the changing position and attitude of the gyro will result in odd and difficult to explain behavior that we observe as anomalies or phenomena. (Very few people have witnessed this phenomena but Sandy has been kind enough to present it to us in this forum.) The explanation of what happens can become convoluted and itself difficult to understand.
The simplest way (that I know of) to achieve a basic understanding is to realize that radial mounted gyros create a TUG-OF-WAR between centripetal acceleration/force (i.e. centrifugal effect) and the precession that results from the angular turning of torque.
It may be initially unclear how to separate these 2 aspects of force that appear to be the same in spinning / turning objects.
The best way I know how to differentiate the 2 forces is; the torque which turns the wheel is the causal force, while the centripetal/centrifugal forces are the result of the turning action (one is the cause of turning the other one is caused by and a result from the turning - i.e. torque causes turning, and centripetal/centrifugal results from turning).
Now that we have a bit of a distinction it may be easier to see how these 2 forces may work in unison during some stages and against each other during other stages, depending on which direction the axis of gyro is pointing (in radial mounted gyros).
It appears intuitively evident that anything that inconsistently fights against itself (one force against another), can NOT be efficient and can only be MARGINALLY effective.
The complete solution may be self evident to those who have sufficient understanding of the dynamics, and who can understand my style of writing. (Those interested in the full explanation please post your request and email adders in this thread and I will contact you.)
Thank you, Luis |
| Date: |
21 January 2006
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Answers (Ordered by Date)
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| Answer: |
Sandy Kidd - 27/01/2006 08:08:39
| | | Dear Luis,
In the future I may very well be proved wrong, but as far as all my many experiments have shown, independent manipulation of a gyroscopic system on its own will not produce inertial thrust.
As I think that trying to develop passive systems to generate inertial thrust are a complete waste of time, my comments here relate to accelerated systems.
In every instance where I have generated inertial thrust it has come about, utilising the association/interference, of forces outside the gyroscopic system.
I spent years, too many years trying to manipulate gyroscopic systems in order to produce some kind of differential which would offer inertial thrust.
I honestly think that at the end of the day I have built and tested every conceivable permutation of gyroscopic rotation speed, system rotation speed and geometric manipulation of the gyroscopic system, and this has been quite some effort I assure you.
The net result is as I previously suggested that Newton will be satisfied unless other forces are allowed to operate along with the system.
I do however realise that enthusiasts will continue to try to find a differential within the gyroscopic system, that being their choice. I did not get one, which has cost me many years of my life, and If I carry on like this I’ll have myself crying.
No pity please I am just a very stubborn individual with unlimited patience.
However in saying all this, here is but one method of producing inertial thrust.
If the geometric change/s, is/are made whilst the considerable system driving torque is removed (temporarily of course, or cyclically) continuous inertial drive can be generated. There would be no fun in it if I told you anymore.
The machine may appear heavier or lighter depending on how the action is created.
I am sure you can figure out a method of doing this.
Will give you something to think about this weekend.
Best of luck,
Sandy
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| Answer: |
Jerry Volland - 27/01/2006 18:08:06
| | | I've had considerable success generating thrust, including extended duration levitation, using split gyroscopes with phase shifted precession. With a split gyroscope, the counter rotating weights on seperate shafts produce gyroscopical reaction forces in the same direction when the unit is tipped. So the THRUST iitself is a REACTION to the tip. Also, in an accelerated system, both the rate of the gyro's rotation and that of the tip are increased in one direction, but decreased in the other. (But only after the phase angle has changed.) So, the tip radius is shortened as the rotation increases. -JV
examples: www.spaceoffice.us/ipm.htm
email: jerryvolland@spaceoffice.us
P.S.: I wish everyone would post their emails, so they can be notified of a responce.
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| Answer: |
Luis Gonzalez - 30/01/2006 00:46:04
| | | Jerry,
I looked for a levitated device on your website but could not find one.
I also read through, looking for an explanation of device or experiment that showed levitation but again failed to find one.
Is the meaning of your term “levitation” something other than an object that hangs in mid air for a period significantly longer then great basketball players are able to remain off the ground?
If your use of the term levitation means something that remains the same distance off of the ground (without aerodynamic help) for a measurable period of time, please tell us how long did the “levitation” last!
Much more important, please put pictures in your website (and eventually videos) of the device as it is in a levitated state.
There is one not often found and unquestionable fact that I did find in your website, where it seems to state that a special configuration with a wheel that is increasing in speed is able to produce a force that stops when the wheel’s speed becomes constant. In my opinion this is a key principle in producing the type of propulsion that we seek.
I look forward to your explanation of “levitation” and I hope it means what we have all been waiting for (backed by pictures, videos, and soon your picture in the news as the first greatest inventor of this century).
Of course if we have different definitions for the concept, than we will all be happy to still have an opportunity at the honor of being the first to builds a machine that levitates through an entirely internal mechanism.
Thank you,
Luis
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| Answer: |
Jerry Volland - 30/01/2006 16:27:38
| | | Louis:
I have a considerable amount of text on my other site:
http://www.freewebs.com/attatchments
A picture alone isn't going to work, because someone will say that I just dropped it. You wouldn't believe the amount of flack I've gotten over this. (For instance, the pendulum thruster tipping the "A" frame over, "doesn't prove thrust, it just destroys the reference frame".) I've put a video capture card on my computor, and am trying to learn how to use it. I actually have two devices I want to make AVI's of - the main machine and the airborne thruster.
My term "levitation" refers to the fact that the device remains in the air, even if not at the same height, for the extended duration of 3 to 4 seconds. Additional building can most certainly, in my mind, extend the flight time indeffinately.
The breakthrough came when I discovered how to generate mechanical quadrature (rotation around two different axies), according to the Gauge Theory of Gravitation, by tipping a Split Gyroscope (counter rotating weights) with phase shifted rotation. This produces thrust in any of the four directions which are perpendicular to gravity. I've had a Split Gyroscope for sale on the Internet in 2 different places for the last couple of years but haven't recieved any interest.
When the two weights are counter rotating they only produce centrifugal force through two 120 degree sectors, one in each direction. When the weights are lined up, or nearly so, before the centrifugal force appears in one direction or the other, torque is produced momentarilly as the machine tips backwards around its base, relative to the weights, in reaction to the weights being pushed out at the top of the shafts. This torque is a rotation, rather than a downwards movement. But when the weights have passed the 30 degree angle away from center, the torque dissappears and centrifugal force starts up, tipping the machine back from the extent of its reverse torque rotation, and beyond. This causes the weights to rotate around their shafts AND around some point in line with, but beneath the bottom of the shafts. As the plane of the weights' rotation is tipped downwards, the gyroscopical reaction is in the same direction on both sides - with the phase shifted precession. However, if the machine were only tipped back and forth as the weights went around - rather than allowing the weights to move out first - the tipping would only cause the weights to speed up and slow down, without any perpendicular thrust.
I've learned that the velocity in the centrifugal acceleration equation (F=V^2/r) applies to a weight's velocity in any direction from which it is tipped, not just to the speed of the tip. That's why a satelite speeding downrange moves upwards in reaction to gravity's slightest downwards tip, maintaining orbit by counteracting gravity.
The bi-axial rotation of the weights consists of a rotation around the top of the shaft at the same time the top of the shaft rotates around a point somewhere beneath the bottom of the shaft. This secondary rotation produces upwards centrifugal acceleration against all mass, even that not associated with the rotors themselves. This rotation is caused by the centrifugal force of the rotors, pulling the top of the device latterally, with the mass dragging below the pull point forcing the curve. The tipping of the wieght's *plane of rotation* (around the top of the shaft) produces centrifugal acceleration (A=V^2/r), which is different than the latteral centrifugal force (F=MV^2/r) the weights also produce. Only on-going acceleration can counteract gravity's acceleration, so each point of mass counteracts its own weight. (The amount of mass doesn't matter, as long as it's made to move with a velocity through a radius. Hence, Einstein's antigravity formula - A=pi*mu*nu [180 degree radius times mass times velocity].) But, experimentally, the acceleration only occurs with the counter rotating weights past the straight-in-line position when the tipping starts. (E.g.: Phase shifted precession.) This is consistent with the Gauge Theory of Gravitation which shows that a graviton is produced by two counter-rotating photons (timewise), with one having a pi/2 phase shift.
Amazingly, while on quadrature, the motor speed lugs down to 15 RPM and the machine moves in slow motion, with the motor groaning loudly and the weights moving through only 120 degrees of rotation, during the 3 to 4 feet of latteral movement before the larger curve intersects the floor, where the direction of lateral movement reverses. And the machine remains perfectly level on flight, side to side, even though the weight of the motor is closer to one side, and, also, even though one end angles up, levels off, then angles down as the machine comes back down. -Jerry
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| Answer: |
Luis Gonzalez - 06/02/2006 01:37:59
| | | Jerry,
I am enjoying your web sites very much and after lengthy reflection I have some questions.
There are 3 sets or types of cycles that jump out from your design.
-The first type of cycle is the take-off hover and fall back to the ground cycle.
-The second cycle is the machine “tipping” cycle.
-The third cycle is the full period of the opposite rotating weights (360 degrees).
1) The hovering, functional results cycle is marked by the take-off, moving horizontally, and then falling back to the ground, which you said takes bout 3 to 4 seconds. Should we add 1 more seconds for the time it takes to raise, or was that included?
2) The machine tips (in one direction and then in another?) in a cycle that appears to be of the same time-length and synchronized with the hovering functional results cycle. Is this correct, or do multiple “tipping” cycles occur during each “hovering” cycle?
3) The weights rotate in opposite direction synchronized to each-other. How many of these full rotations cycles occur for each full “tipping” cycle? Is the rotation of the weights at a constant angular velocity or does it vary depending on what position the tipping cycle is at?
Also in the second sentence of your fifth paragraph above you say <“When the weights are lined up.”>
How do you mean that? Are they in a straight line at furthest position from each other, or nearest position to each other? Or, are they both pointing upward or downward?
Please clarify these questions so that I may start to understand your very intriguing devices.
Thank you, Luis
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| Answer: |
Jerry Volland - 07/02/2006 15:06:16
| | | Jerry
I am enjoying your web sites very much and after lengthy reflection I have some questions.
There are 3 sets or types of cycles that jump out from your design.
-The first type of cycle is the take-off hover and fall back to the ground cycle.
-The second cycle is the machine “tipping” cycle.
-The third cycle is the full period of the opposite rotating weights (360 degrees).
The *momentary* tipping cycle is the period just before lift-off. (This is when the phase shift occurs.) At this point, the lower frame tips up on one end, *until* it forms an angle of 60 degrees with the ground. (This angle, in the opposite direction, is preserved when it comes back down.) With low power, a few complete cycles of operation are required to reach the 60 degree tip, with a little more *momentary* tip added each time the phase shift occurs (with one end touching the floor). Once this angle is reached - when the backwards angle of the frame plus the angle the weights have moved forwards equals 90 degrees - no additional *momentary* tipping occurs with continuing operation. After that, all tipping occurs in the air, as the machine is moving, so "tipping cycle" then basically refers to the moving through the arc in the air, which is an on-going "tip" of the lower frame. Each of these "moving tipping cycles" occurs during 180 degrees of rotation, and are therefor only "half cycles" of the full period of rotation.
1) The hovering, functional results cycle is marked by the take-off, moving horizontally, and then falling back to the ground, which you said takes bout 3 to 4 seconds. Should we add 1 more seconds for the time it takes to raise, or was that included?
The 3 to four seconds is an estimate - without a stopwatch - and does include the time it takes to leave the ground. But the machine does not "fall back" to the ground. It remains in levitation until the leading edge of the frame intersects the ground.
2) The machine tips (in one direction and then in another?) in a cycle that appears to be of the same time-length and synchronized with the hovering functional results cycle. Is this correct, or do multiple “tipping” cycles occur during each “hovering” cycle?
You are correct, the cycles are sequential, first in one direction, then the other. These bi-directional cycles take exactly the same amount of time.
3) The weights rotate in opposite direction synchronized to each-other. How many of these full rotations cycles occur for each full “tipping” cycle? Is the rotation of the weights at a constant angular velocity or does it vary depending on what position the tipping cycle is at?
Only 1/2 full rotation cycle occurs for each ful tipping cycle. The rotation speed stays constant - around 15 RPM.
Also in the second sentence of your fifth paragraph above you say <“When the weights are lined up.”>
How do you mean that? Are they in a straight line at furthest position from each other, or nearest position to each other? Or, are they both pointing upward or downward?
For one complete tipping cycle, the weights begin by being lined up in a straight line at their nearest position to each other. For the next, reverse direction cycle, they start off at the furthest position. Together, the two tipping cycles comprise one full 360 degree rotation. -Jerry
Please clarify these questions so that I may start to understand your very intriguing devices.
Thank you, Luis
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| Answer: |
Luis Gonzalez - 09/02/2006 17:26:02
| | | Jerry,
It sounds like all the three cycles that I mentioned have a period ratio close to 1:1:1 (in relationship to each other) and all three have cycle duration of about 4 seconds (please correct me if I am wrong).
I have a four more questions regarding the device and a couple more on clarification of the writing:
-What is the length of the radius of rotation of the 1lb weights (and the distance between their centers of rotation)?
-What are the dimensions of the machine (LxWxH with and without the extended aluminum square-tubing that has the extra wheels)?
-What is the function of the four springs?
-Important- How long does the device remain completely off the ground (where no portion of it touches the ground)?
-How far off the ground is the device when it is parallel to the ground?
-What do you mean by “*momentary* tipping?”
-How is tipping induced, or is it something that just happens on its own?
-What is it that rotates 180 degrees during each “moving tipping cycle?”
Your device is fascinating as are the connections you claim in relationship to quantum mechanics.
Thank you, Luis
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Luis Gonzalez - 10/02/2006 02:24:04
| | | Jerry,
I forgot to ask, what is the total weight of the device?
Thank you, Luis
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Jerry Volland - 11/02/2006 17:07:06
| | | Jerry,
It sounds like all the three cycles that I mentioned have a period ratio close to 1:1:1 (in relationship to each other) and all three have cycle duration of about 4 seconds (please correct me if I am wrong).
This is not correct; only the movement through the air has an extended duration. For better clarity, the operation can be classified as having two cycles, each with two components. The first component is the tipping up on the end of the base. The second component is the arcing through the air movement. The second cycle is the same as the first, except in the reverse direction. The tipping component always occurs around the end of the base which is opposite to the direction of subsequent travel. When it first starts, this tipping is very slight so the distance of travel through the air, and the height of the arc, is small. However, the angle of tip - before liftoff - is preserved when the device comes back down on the other end of the base, at the end of the through-the-air arc. At this point, the momentary tipping recurs and the tip angle is increased. (This momentary tipping component is quick and lasts for around 1/2 second.) This increase in tip angle each time, and the increase in distance of subsequent travel, continues until the tip angle equals around 60 degrees. At this point, the torque goes straight down against the ground and there's nothing to cause the tip angle to change. After this point, the machine only arcs up, through the air, and back down, then repeats in the other direction, for as long as it's operated. But there's still a split second pause before re-liftoff, as the phase shift changes direction.
I have a four more questions regarding the device and a couple more on clarification of the writing:
-What is the length of the radius of rotation of the 1lb weights (and the distance between their centers of rotation)?
It's 6 1/4 inches from the center of the vertical shaft to the tip of the weight, with 13 5/8 inches between the shafts, center to center.
-What are the dimensions of the machine (LxWxH with and without the extended aluminum square-tubing that has the extra wheels)?
It's 6 1/2" long, without the four springs (13 1/2" to the outsides of the spring brackets) by 18" wide, with the weights pointing straight in (27" with the weights pointing straight out) and 15" high (without a base). The base itself is around 23", but could be around 18" if not off center. The plastic tubing which looks like aluminum didn't work out, since they broke during the first torque impulse. (Phase shift tip.) The best thing I've found for the base is 1"X2" fir strips, with the wide side horizontal. The flat wood is just springy enough that it doesn't break, and this springiness itself may have some effect on the operation. I've found that with centrifugal force, a spring against the shaft or frame will result in a 90 degree phase shift. (So a tether can be replaced with a spring.)
-What is the function of the four springs?
The springs are for non-quadrature operation. In this mode, the machine operates as a true Split Gyroscope and only the vertical drive shafts are allowed to tip, rather than the base of the machine. This happens because the bolts holding the upper bearings are loose, so the bearings can slide back and forth, allowing the tops of the shafts to move relative to their bottoms. (For quadrature operation, these bearing bolts are tight, the shafts can't tip relative to the base and the four springs are irrelevant.) Without quadrature, the RPM's stay constant at 600 and the machine jumps straight up - around 9" - rather than arcing. I'm still developing the system and have only tested it with two springs, both on the same end. During this test, the base (without wheels) slid on the grass a short distance, dug in and stopped, then the machine jumped straight up into the air - proving the shaft tipping effect. Without the other two springs, the machine then started moving horizontally in the air, but immediately and very quickly torqued the leading edge of the base down at an angle, then moved down at that angle and hit the ground. When the base dug in and stopped, the tops of the shafts continued moving, stretching the two springs on the back side. Also, the weights continued turning - with an off center rotation - and this caused the required phase shift, since the weights were ahead of center when the springs pulled the shafts back to vertical. An off center rotation is equilalent to torque and does not produce centrifugal acceleration (A=V^2/r). However, the weight's moving ahead of center also has the effect of lengthening their tip radius - relative to the bottoms of the shafts - when the tops of the shafts are pulled back towards the vertical. And the length of the tip radius is important to the production of centrifugal acceleration, which is a gyroscopic force.
-Important- How long does the device remain completely off the ground (where no portion of it touches the ground)?
When moving on quadrature, the time period in which no portion is touching the ground is around 4 seconds.
-How far off the ground is the device when it is parallel to the ground?
Around 12 to 14 inches. Interestingly enough, at this point the spokes holding the one pound weights are pointing in the direction of travel, perpendicular to the straight-in-line positions. Quadrature operation produces greater lift that just the shafts tipping, since the arcing produces centrifugal acceleration, and lift, for every point of mass in the machine. However, the non-quadrature operation may be superior, since the motor speed doesn't lug down and the lift cycles can be repeated much more quickly. (600RPM vs 15 RPM.) This is true even though tipping the shafts alone only produces centrifugal acceleration against the weights and their spokes.
-What do you mean by “*momentary* tipping?”
The tipping component of the operation lasts for less than a second, contrasting with the *extended duration* of the time it spends off the ground.
-How is tipping induced, or is it something that just happens on its own?
The wheels are only allowed to roll a short distance (less than 2") before hitting an obstruction, such as small rocks. The effect can be demonstrated by spinning a small top in such a way that it "walks" until it bounces off an obstruction. This completely changes the type of spin the top exhibits. Stopping the horizontal movement of the base frame early causes the tops of the shafts to continue moving, relative to the end of the base. This movement of the tops of the shafts relative to the end of the base is what causes the tipping associated with the phase shift, with or without the four springs. Changing the type of spin is what triggers the quadrature effect. But there is some automatic effect which just happens on it own. When the device is mounted with a journal bearing centered on each side and stiff springs at the bottom on each end, the top of the machine tips twice in the same direction during each rotation, once as the centrifugal force on one side pulls the top in that direction, then again, in the same direction, as the weights are lined straight up, starting to move out on the other side.
-What is it that rotates 180 degrees during each “moving tipping cycle?”
The spokes holding the weights and the vertical drive shafts. At the start of the first cycle (just prior to the first movement through the air), the weights are lined straight up next to each other in the middle. By the time the machine comes down on the far end of its base, the weights have moved around until they are lined straight up at the furtherest position away from each other. During the next cycle of movement through the air (in the opposite direction) the weights continue moving on around until they are once again lined straight up in the middle by the time the machine comes down on that end.
Your device is fascinating as are the connections you claim in relationship to quantum mechanics.
Thank you, Luis
I'm glad you like it. I've open sourced it, so you and anyone else can have it. Some bright person will figure out how to force the phase shift using a vertical shaft tipping cam (and the four springs), or, put a couple of these machines in a horizontal track so they can slide towards and away from each other The horizontal track will then absorb the torque impulses, stressing that sub-frame without producing any downwards movement. Building a better Space Drive has always been on my Agenda. I discovered this Mechanical Levitation back in '94, but I've since found something else which is a lot better.
I forgot to ask, what is the total weight of the device?
Thank you, Luis
It weighs around 35 pounds. -Jerry
"If God wanted Mankind to have Mechanical Levitation, it would be in the Bible. (Ezekial 10:10)"
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| Answer: |
Luis Gonzalez - 12/02/2006 18:44:24
| | | Jerry,
The force (F=MA) of earth’s gravity holding a 35lb device down is (35lb)*(32ft/sec-sec) = 1,120ft-lb/sec-sec.
The total acceleration of your design comes from two 1lb weights, each one rotating at 15RPM = 0.25CPS with a radius of 6.25 inches = 25/48ft.
From these dimensions we can calculate a velocity of about 0.82ft/sec.
(Assuming that the tipping is somehow induced by the motor, we may add the velocity of tipping over a max 120degres arc per jump at about 1/12CPS with a radius that can not exceed 17 or 18 inches = 1.5ft.
From these dimensions we can calculate a velocity of about 0.7854ft/sec.)
Adding both these velocities (0.82ft/sec) + (0.7854ft/sec) = (1.61ft/sec) total max velocity.
Centripetal/Centrifugal force:
= (M)(V^2)/(R)
= (2)(1.61)(1.61)/(25/48)
= 9.953664ft-lb/sec-sec (say 10ft-lb/sec-sec)
We need a force a bit over 1,120ft-lb/sec-sec to separate the device from the ground.
The dynamics of the device are producing less than 10ft-lb/sec-sec.
I can’t see where the additional momentum is coming from to create sufficient lift.
Please tell me where I have calculated wrong or what I am failing to see.
Thank you,
Luis
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| Answer: |
Jerry Volland - 13/02/2006 13:58:15
| | | The Centrifugal Acceleration (A=V^2/r) has nothing to do with the velocity of the weights around the tops of the shafts. Don't add this to the arcing velocity - this is what causes the arcing velocity.
Also, F=MA has nothing to do with it, except to calculate the speed of the arcing movement.
The acceleration of gravity holding a mass down is g. This is true no matter how much mass there is. In order to produce levitation, the UPWARDS ACCELERATION has to be equal to or greater than g. With Centrifugal Acceleration, the amount of mass doesn't matter, as long as we know it's moving with a Velocity V. (This is equivalent to the speed of the arcing movement.) This is true no matter how much mass there is.
Your figure for the radius of the arc (1 1/2') seems pretty close, so square the Velocity and divide by this number, to see how the estimated upwards acceleration compares with that of gravity.
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Jerry Volland - 13/02/2006 14:49:11
| | | On second thought, the velocity of the weights around their shafts would add to that of the arcing movement, but only as long as the weights are moving in that direction. Also, the arc radius of these weights is greater than that of the various points throughout the rest of the machine. However, this compound velocity applies to the Centrifugal Acceleration of these weights only, since the rest of the machine has only the arcing velocity.
Also, when using your presented equation, the mass is 35 pounds and the radius is 3/2'.
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Luis Gonzalez - 17/02/2006 12:19:45
| | | Jerry,
I want to say this in a nice way. You have made incorrect statements. I will correct the more basic errors, only because this is my thread (a basic physics course or book would work as well). I will start by stating basics.
1) FACT - Mere mortals can not create acceleration by itself without using a mass. Something needs to be accelerated (does anyone know something other than a mass that can be accelerated?).
2) IMPORTANT - The force exerted upon a mass always creates an acceleration (F=MA) or (A=F/M). (OK?)
3) The force (F) required to lift a one-pound-weight is F=MA = (1lb)*(32ft/sec-sec); that is sufficient to accelerate 1lb to 32ft/sec-sec, just enough to match gravity’s acceleration (A=g).
4) 1,000 pounds requires 1,000 times the force required to lift one pound.
F=MA
F1=1*32 = 32
F1000= 1,000*32=32,000
The gravitational acceleration is constant, but NOT the gravitational FORCE!
We can’t use the motor that flies a toy plane to fly a cargo plane!
A force of proportionate magnitude is required to lift a mass of proportionate magnitude.
Though the forces needed to lift (or hover) different masses MUST be of different magnitude, the resulting acceleration for all the masses are the same amount, 32ft/sec-sec (Get it?).
To see this clearly take the equation F=MA and solve for acceleration giving A=F/M.
Now you can see that to produce a given constant acceleration (such as A=g=32ft/sec-sec) requires a force proportionate to the magnitude of the mass (i.e. need a larger force to give a larger mass the same acceleration).
Most people have a problem seeing that a force is applied to yield acceleration, NOT the other way around. In other words, you can not apply an acceleration to get a force. You can apply a force that produces acceleration upon a given object and the acceleration results in increasing the object’s velocity; this is the correct sequence. Can you see this? So, NO you can NOT just apply the same constant “g” to all different sizes of mass and get lift or hovering!
Part of the confusion results from our knowledge that gravity “g” is constant; however though gravity’s acceleration is constant the force that “g” produces on different amounts of mass is different; that’s why some things weigh more than others.
To address another statement, Centripetal/Centrifugal force does have plenty to do with the velocity of the weights around the top because Centripetal acceleration results from the velocity that the motor drives the weights, and that is the full and only measure of how much energy is being input into the system (there is no other source of energy input into the system, is there?). Where do you suppose that the energy that can cause any motion (e.g. arcing) may be coming from?
The main point is that lifting 35lbs off the ground requires a FORCE (not an acceleration) and the magnitude of that force is 35*g=1,120ft-lb/sec-sec (I did not make this up).
The big problem is that the device in question is not inputting even 1/% of the energy needed.
I think the important items are covered. Please take the time to learn and understand the physics; you will be able to find the other errors on your own.
Thank you, Luis
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| Answer: |
Jerry Volland - 18/02/2006 15:41:30
| | | Luis:
In order to better understand the operation of the device, I placed a bi-fold hinge on each side, on the bottom. (This was in 1994.) This way, when it tips backwards, the top board will open on one end, and when it tips the other way, the second board will open on the other end. I then set the weights in the center, where they were lined up. As expected, when the weights started moving out the machine tipped backwards. Then, after the torque part of the cycle, it settled back down, with the motor making the ususal loud groaning sound. However, there was no upwards movement.
When I first built the machine, I was using large bicycle sprokets that I had painstakingly converted into cog wheels (which were extremely noisy). After I found out that it was producing a usable effect, and still unable to afford gears, I scrounged up the money to buy the smaller, regular sprockets I'm still using. Changing from the cog wheels to the smaller pseudo 'gears' required the entire frame to be rebuilt. To my dismay, it no longer lifted off. After working most of the day changing the frame a number of times, I finally set the length of the bottom runners on the back to around 5" from the edge. Then when I turned it on, it made the tell tale groaning sound and I knew I had it even before it lifted off.
Remembering this difficulty, I rebuilt the bi-fold hinges so that they had this length. This moved the tip point back from the back edge on the bottom. Then when I turned it on, when it came to the point where it had settled back down, it lifted up instead, opening the top hinge. Then, when it reached the other half cycle, the front hinge opened, and it lifted a little higher. (The top board - i.e., the first hinge - didn't close during this part of the operation.) When it repeated the first half cycle, the first hinge opened a little more, and it lifted higher again. Etc.
Through the years, as I experimented with this device in a number of ways, I noticed that the pillow block bearings had a certain amount of springiness, due to their rubber cushions. This springiness allows the tops of the shafts to be pulled back and forth around 1/2" in either direction. Back in '91, I capitalized on this effect by adding two of the four springs and loosening the bolts on the upper bearings. This allowd a couple of inches of movement backwards, but none forwards. However, this amount of movement of the tops of the shafts, relative to the tip point at the lower bearings (with a tip radius of 10 1/2") still produced liftoff, which was quite rapid. Since the original configuration, when modified, didn't produce liftoff until the base boards were set at the right shortness, it's become apparent that the springiness of these baords added to that of the pillow block cushions to allow enough torque storage to function in the desired manner. This torque storage, when released, adds appreaciably to the momentary velocity of the weights around their tip point, five inches back from the edge, on the bottom (with a tip radius of 17 1/2").
However, this still doesn't resolve the issue of the loud groaning sound the motor makes when moving in slow motion on quadrature. This sound, and the slow motion, is absent when the four springs are used to apply stored torque to the shafts. Another factor to consider is that with the bi-fold hinges opening at the edge on the bottom, the sound was present even though the angle of the weights prevented their centrifugal force from adding to gravity's force as it caused the device to settle back down. (There was no horizontal movement.) So it seems indisputable that gravity is coupling with the device while it operates in quadrature mode. Further evidence of this fact is that the force applied by gravity was less when the tip point was at the bottom edge than it was when the tip point was moved back five inches. This is because when the tip point is at the edge, some of the machine's mass is behind the tip point, whereas with the extended tip point, all of the mass is on the same side of this point.
So, some of the force which produces lift off (on quadrature) is supplied by gravity pulling the device downwards around the tip point, in the moment just before lift off. Interestingly, once it starts, this coupling with gravity continues even after the machine is in the air. So the true tip radius, which varies as the weights move out and back, has to be calculated around this tip point at the end of the lower runners, even when the machine is arcing through the air.
Jerry
Centrifugal Acelleration, A, equals V^2/r. For lift off , A must equal g (gravity's acceleration).
32 ft/sec/sec=V^2/(10.5/12)
(.875)*32=V^2
28=V^2
5.29ft/sec/sec=V
32ft/sec/sec=V^2/(17.5/12)
(1.46)*32=V^2
46.7=V^2
6.83ft/sec/sec=V
bi-fold hinge:
________
O_______
_______O_______
P.S.: Go back to your physics book and see how to calculate ORBITAL VELOCITY. Also, how much mass is a SLUG?
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Jerry Volland - 18/02/2006 16:10:08
| | | Luis:
I see that I just made a mistake with my square root units.
So, with a tip radius of 10.5", the required velocity is 5.29ft/sec (rather than sec^2, which would still be an acceleration). Likewise, with the tip radius of 17.5", the velocity is 6.83 ft/sec.
Also, you said:
Most people have a problem seeing that a force is applied to yield acceleration, NOT the other way around. In other words, you can not apply an acceleration to get a force.
Then you said:
Part of the confusion results from our knowledge that gravity “g” is constant; however though gravity’s acceleration is constant the force that “g” produces on different amounts of mass is different; that’s why some things weigh more than others.
The fact is, the acceleration of gravity does produce a force, based upon how much mass is acted against.
Jerry
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Luis Gonzalez - 20/02/2006 00:24:09
| | | Jerry,
Your explanation is not correct. Using the force of gravity to overcome itself is tantamount to building a perpetual motion machine. I value my time very much; please take that line of discussion to others, thank you. I prefer to spend time developing something worth while.
The meaning of SLUG and ORBITAL VELOCITY are known but both concepts often cause confusion to the unprepared (please see short definitions below). The introduction of irrelevant FACTS can only be construed as a distraction from the real issues. Irrelevant detailed verbosity is also a common way of distracting from real issues.
Based on the facts, credibility is low to non-existent. Immediate action pictures may make some people stop and think twice. Picture will be more credible if they include a human nearby and if they are provided as early as possible (it is possible to manufacture pictures but takes some effort).
It is also extremely simple to put short videos in the web; therefore good videos in short notice provide compelling proof but should include a running timepiece. Excuses for not having videos can only further damage all hopes for credibility.
The absence of action pictures and videos speaks volumes.
A man who is clever enough to make a 35lb machine hover for 4seconds at a time (over and over), should be clever enough to contact the local news etc for demonstrations (especially if he now has something even better than the 1994 design!). When you are in the 10 o’clock news, I will wish I had asked to join your team.
I would wish you luck, but you either don’t need it, or it won’t be of much help.
-SLUG - a, not often used, unit that measures mass so that a 1lb force produces a 1ft/sec-sec acceleration. Changing units of measure do not address or add meaning to the issues.
-ORBITAL VELOCITY - depends on the distance from the planet (center) and on the mass of the planet.
Both mass creates an attractant FORCE, NOT ACCELERATION, that interact in proportion to their magnitudes (F=G*M1*M2/d*d). The force produces acceleration upon the masses (NOTE which is the CAUSE and which is the EFFECT). There is no such thing as a naked acceleration without mass!
Thank you, Luis
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Jerry Volland - 20/02/2006 10:27:53
| | | Luis:
If a SLUG equals a unit of mass which equals 32 pounds, why does one pound of thrust lift one SLUG (32 lb) of mass?
Orbital velocity is calculated with the equation:
A=V^2/r
Why is this velocity (at the same altitude) the same for a satelite weighing 1,000 pounds as it is for a satelite which weighs 1,000,000 pounds?
The reason I don't have video equipment is poverty.
Thank you for you time and consideration. I've had enough flack for a while.
Jerry
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Luis Gonzalez - 21/02/2006 00:39:08
| | | Jerry,
Yes A=V^2/R is used to calculate orbital velocity because it is the equation for centripetal acceleration.
The velocity of two satellites weighing 1,000lb and1,000,000lb is the same because they both need to have the same centripetal acceleration (A=V^2/R) to balance against the acceleration of gravity.
However, the larger satellite carries 1,000 times the force of the smaller one; the force is proportionate to the mass.
Even more important, the force needed to bring the more massive satellite to that altitude and velocity is also 1,000 times greater than the force required for the smaller mass.
Again, (as explained a couple of times before) it is a matter of cause and effect!
A greater force was needed to get the heavier unit up and to the same velocity. Do you see?
Don’t skip the part where the satellite gets put into orbit because that is the part that is applicable to our devices.
Once the device is up there, it has the necessary impulse to continue along regardless of the mass, but you must get them there first. Do you get it now?
What’s the reason your devices are not in the news and made you a fortune?
Being poor isn’t wrong but pretending to have something you don’t is wrong and always causes a lot of flack.
Also, being poor doesn’t mean that I have to be fooled by your erroneous statements or even to pretend that I accept them. Especially when I am trying to accomplish something that I think is very important.
If you really have a device that you claim then you will soon be very rich (I will not), and all the flack in the world should not mater.
If your device does not work as claimed then you have brought the flack upon yourself.
I am sorry, but you are much too industrious to let words stop you from making progress.
If you have what you claim you would borrow video equipment, or go to a junior college and take a course where you can use the equipment.
Whatever the situation is, I wish you luck Jerry Voland.
Sincerely, Luis
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Jerry Volland - 21/02/2006 12:33:29
| | | Luis:
Thank you for the warm wishes, and the suggestion about the Junior College. I just may do that. Right now, I got a little digital camera for $15 that makes AVI's. It only takes 2 frames per second, but that might still show something.
As far as the amount of force needed to achieve liftoff, did you go back and redo your calculations based upon the amount of force stored, and then released, by the springs? In the case of the Split Gyro application (the four springs, with the lower bearing as the tip point) this force would seem to be substantial. The phase shift sequence occurs in 1/120 sec (maybe a little less - but only a little). During this interval, the tops of the shafts are pulled back around two inches. I don't have the focus, or the time, at this point to calculate how fast the springs snap back. (Maybe you do?) Once the true velocity of the weights around their tip points is known, all that is needed is to apply the centripetal acceleration equation to see if the acceleration produced equals or exceeds that of gravity.
And I may very well go on the news, but I'm extremely reluctant to do so. I've already experienced being recognized by everyone (in a different town), and I have no desire to walk that road again.
Jerry
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Jerry Volland - 21/02/2006 12:54:58
| | | P.S.: I don't expect to make any money with this. I've systematically placed it into the Public Domain over the past five years, and open sourced as many details as I, or anyone else, could think of. (But you're the first one who did any calculating.) Anyone who wants to can use this (including industry), without paying me anything. Anyone out there who "gets it" can now have this. But if the world waits for me to get enough money to patent it, etc, then no one will have it.
Besides, I've found something electrical which is a lot better. (I am an electrician, after all.) Once Private Space Industry takes off, investor interest will be in my favor.
The main point I've been trying to show is that those who want Inertial Propulsion should stop thinking "Centrifugal Force" and look at Centrifugal Acceleration. IOW, think outside of the plane. I have a love for the Quest myself, and am just trying to help.
Jerry
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Luis Gonzalez - 23/02/2006 03:09:36
| | | Sandy,
Sorry I have been otherwise occupied and have not had time to respond.
We are in agreement on two major things.
First, I intuitively have never allowed myself to fantasize (except maybe when I was 12 years old) that a passive system could generate the mechanical propulsion that we seek (assuming I understand what you mean by a passive system).
Second, I also agree that force(s) external to the gyro system must come into play to obtain propulsion. All of my (privately documented) designs use interactions that also include forces originating external from the gyro and from the rotation system.
We are both aware that the accelerated design that I described in the introduction to this thread will not yield much propulsion (not at the stage that I stopped describing it).
You know it based on results from your numerous experiments, and I know it based on results derived from analysis of the forces involved and the fact that the 2 major forces work largely against each other.
I am certain, Sandy that you have derived a solution. I also have derived a solution through my analysis.
So, we have either the same or completely different solutions to the same or completely different problems.
Neither of us would expect the other to divulge our solutions at this stage in the game (at least until we have built and demonstrated an undisputable proof).
I am going to share the relational equation that I use to derive approximations of how much acceleration the flywheel(s) in my design solution will generate in the direction of precession (hopefully in an upward direction).
The result from this equation is a magnitude of acceleration that result from the ratio of system-radius (R) to flywheel-radius (r) and the ratio of system-period (P) to flywheel period (p), using my design solution.
By examining this equation you may be able to manipulate the terms backwards and figure the origin of the equations that I used to derive my solution. If you get there it will be a short jump to figure out what my solution design is.
If you manage this clever piece of work you will also be able to determine whether our solutions are the same or not (that is if you have an equation to calculate the magnitude of force that your design is expected to generate).
This is my equation A= [(4)(PI)(PI) x (R)(R)(R)(R) x (P)(P)(P)(P)] / [(r)(r) x (p)(p)]
Which may be written in different notation as A = (4PI^2)(R^4)(P^4) / (r^2)(p^2)
Where:
A = Acceleration generated per pound of flywheel mass
PI = 3.1415…
R = Radius of the system
r = radius of the flywheel
P = Period of the system in full turns per second (CPS)
p = period of the flywheel in full turns per second (cps)
In English:
Acceleration = 4 times PI squared, time the Radius of System to the fourth power, times the Period of the system to the fourth power; divided by the radius of the flywheel squared, times the period of the flywheel squared.
To obtain the force just multiply this acceleration times the mass of the flywheel.
The results predicted by the equation derived from my design are encouraging.
With:
System radius = R = 2ft
Flywheel radius = r = 0.25ft
System period = P = 2cps
Flywheel period = p = 2cps
The predicted resulting acceleration is 16,384ft/sec-sec
This acceleration is ON for over one third of the full cycle generating an average acceleration of over 5,461ft/sec-sec for each flywheel.
Flywheels with acceleration of this magnitude should be able to work against a full device mass that is up to 170 times greater than the mass of a flywheel.
You may be thinking that such acceleration will tear any of your designs apart. If that’s the case, our designs can not be too similar because mine flow smoothly with one exception, when the momentum of the flywheel(s) is/are absorbed by the device.
I have 2 questions for you Sandy:
1) Do you have an equation or a way to calculate the force that your device will generate?
2) Are you willing to share your equation or algorithm?
Thank you, Luis
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Elder Dr. & Mrs. Hsien-LuHuang - 23/02/2006 15:55:51
| | | Dear Mr. Luis Gonzalez:
Thanks for the detailed presentation of your analysis quoted below. However, we like to bring up two points for your reexamination of your formula and your numerical calculation. Thanks.
I.
The dimension of the left hand of side your equation is L/(TT), but
the dimension of the right hand side of your equation is [(LLLL)/(TTTT)]/[(LL)/(TT)], which is (LL)/(TT). Therefore,
The dimension of the left hand side of your equation is not equal to
the dimension of the right hand side of your equation.
II.
Substituting the numerical figures given by you into your equation, we have
[(4x3.1416x3.1416)x(2ftx2ftx2ftx2ft)x(2/sec)x(2/sec)x(2/sec)x(2/sec)] / [(0.25ftx0.25ft)x(2/sec)x(2/sec)] =
[40x16x(ft)(ft)(ft)(ft)x16x(1/sec)(1/sec)(1/sec)(1/sec)] / [(1/16)x(ft)(ft)x4x(1/sec)(1/sec)] =
40x16x16(ft)(ft)x4(1/sec)(1/sec) =
160x16x16 (ft)(ft)(1/sec)(1/sec) =
40960 ft ft /(sec)(sec) which is not equal to 5,461 ft / sec-sec as calculated by you.
Thanks for your attention.
God bless you.
Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang
P.S. Listed below for your reference is the text quoted from Mr. Luis Gonzalez dated 2/23/2006
This is my equation A= [(4)(PI)(PI) x (R)(R)(R)(R) x (P)(P)(P)(P)] / [(r)(r) x (p)(p)]
Which may be written in different notation as A = (4PI^2)(R^4)(P^4) / (r^2)(p^2)
Where:
A = Acceleration generated per pound of flywheel mass
PI = 3.1415…
R = Radius of the system
r = radius of the flywheel
P = Period of the system in full turns per second (CPS)
p = period of the flywheel in full turns per second (cps)
In English:
Acceleration = 4 times PI squared, time the Radius of System to the fourth power, times the Period of the system to the fourth power; divided by the radius of the flywheel squared, times the period of the flywheel squared.
To obtain the force just multiply this acceleration times the mass of the flywheel.
The results predicted by the equation derived from my design are encouraging.
With:
System radius = R = 2ft
Flywheel radius = r = 0.25ft
System period = P = 2cps
Flywheel period = p = 2cps
The predicted resulting acceleration is 16,384ft/sec-sec
This acceleration is ON for over one third of the full cycle generating an average acceleration of over 5,461ft/sec-sec for each flywheel.
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Luis Gonzalez - 26/02/2006 14:25:53
| | | Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang
Thank you for examining my presentation and for catching my errors. (It looks like I have made a mess of things.)
My first equation represents neither acceleration nor kinetic energy, though it resembles the latter for a reason (sorry).
After serious reflection and work I have derived an equation that I feel more comfortable with.
The new (more accurate) equation that represents acceleration for my design is:
A = [(PI)(R^2)(P^3)] / [(r)(p)]
When fed the original distances and rates (R=2; P=2, p=2 & r=0.25), this new equation yields an acceleration of only 201ft/sec-sec. This number is not as impressive as 40960 or any of the previous numbers.
However, a projected acceleration of 201ft/sec-sec indicates that each pound of flywheel will potentially lift 6 times its own weight (under earth’s gravity).
Further, the acceleration produced increases in proportion to the cube of the period of the system (the current yield of 201 assumes a period of only 2cps).
I apologize for the misunderstandings that my errors may have caused. It’s easy to cause the trouble but not as easy to sort it out. (Hast and eagerness, what waste and trouble thou caused.) The best we can do is to find the error, admit it, repair it, and thank those who helped to fix it.
Thank you, Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang, for your help.
(Please let me know if you see any more errors)
Regards,
Luis
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Luis Gonzalez - 01/03/2006 02:57:58
| | | Sandy,
Fixing the errors has sort of clouded my questions to you.
Now that the errors sorted, will I hear a response from you?
Thank you, Luis
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Sandy Kidd - 20/03/2006 08:37:21
| | | Dear Luis,
Please accept my apologies for the delay in replying to you.
I found considerable difficulty in answering your last posting.
This is my second attempt at it.
I do have working formulae to base my machine designs on, but they are somewhat different in direction to the one you are taking.
From years of zero results, some mediocre results, and a few good results, I found it necessary to reappraise the whole situation, which eventually resulted in a step change in philosophy and consequently machine design.
Once I discovered how and why a gyroscopic system needed to react, to create the conversion, the priority returned back to the gyroscope.
.I get the impression that your formulae are designed to calculate the vertical acceleration of a gyroscope in an accelerated system.
I agree with what you are aiming to do but I really cannot see a gyroscope changing speed fast enough to produce a sine wave output, unless a very, very flattened one, with the attendant efficiency losses.
Let me put it another way. I cannot visualise a gyroscope operating within the total envelope and being able to complete one complete speed change cycle within several revolutions of the system.
From my own experiments Luis, I observed that as the gyroscope rotation speed is increased the angular momentum in the system diminishes in direct proportion to the speed increase.
In light of this the gyroscope is not forced to overcome massive amounts of inertial mass. It transfers its own mass proportionally as its rotation speed is increased, to the point where it has no mass left to accelerate.
This is the “mass transfer” effect that Eric Laithwaite found so hard to get to grips with.(see “Heretics”) Why it appears to be so hard to measure beats the hell out of me? It is easily seen and easily measured.
Similarly if the gyroscope speed is constant and the system is accelerated the result is the same. The system load diminishes on acceleration and increases on braking
Unfortunately gyroscopes are like that, every normal prediction is reversed.
Not that it matters but I am led to believe that Euler’s Equations predict the angular momentum loss, but not being up on this complicated mathematical stuff I do not know far they go, or what inspired the predictions.
However it is in this particular area that the breach with Newton lies, so I tend to believe that Euler’s Equations were limited in their extent, or this discussion would have taken place some considerable time ago.
My three score and ten comes up next year so I have left it just a little bit late to get involved in this stuff
Would it be at all possible to prevail upon Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang to help out with this one?.
In the final analysis I am of the considered opinion that pure gyroscopes are just not capable of producing what is required to do the job. They can be made to produce significant amounts of thrust but due to the gyroscope’s own inertia, switching rotation speeds is limited, and this of course chokes the output to a small percentage of what is available in the operational envelope.
I have steered away from associated gyroscopic speed and system geometric changes for reasons of engineering complication and vibration and have concentrated for the past few years, on alternative methods of producing differential
This has turned out to be a bit easier than I envisaged. Now a vertically, and horizontally, thrusting gyroscopic device (same device) is possible and testing of such a device is imminent.
I’ll call it a day when I see a good result on a ballistic pendulum test.
I’m hoping this is sooner rather than later.
Regards,
Sandy.
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Elder Dr. & Mrs. Hsien-Lu and Hui-Lien Peng Huang - 20/03/2006 23:06:33
| | | Dear Mr. Sandy Kidd:
Thanks for mentioning our names in your recent response to Mr. Luis Gonzalez.
By the grace of our Lord Jesus Christ, don’t worry about your age. Our Lord Jesus Christ let Elder Dr. Hsien-Lu Huang, now 83, pursue his advanced degrees in Electrical & Electronics Engineering in Virginia Tech when he was 43 years old and finish his M.S. and Ph.D. in three years when he was 46, although he spent all his Friday nights in leading a Bible Study Fellowship on the Campus, and his Sunday mornings in teaching adults Sunday School Classes in a Church in Blacksburg, Virginia then. Our Lord let Elder Huang, now age 83, keep on working in the United Space Alliance to support the Space Operations of NASA Johnson Space Center since 1979, after being an E.E. professor in West Virginia University, and years of working in industries in Chicago, Ill. and Horsham, Pa.
Below we would like to share with you our Bible study fellowship with our grandchildren on Prov. 1:7, also Prov. 9:10, and Psalms 111:10, and how to fear the LORD learned from Prov. 2:1-5.
God bless you, your family, your research, your invention, and your work.
Elder Dr. & Mrs. Hsien-Lu and Hui-Lien Peng Huang
AD 2006 NEW YEAR MESSAGE TO GRANDCHILDREN
ELDER & MRS. HSIEN-LU & HUI-LIEN PENG HUANG
JANUARY 14, 2006, SATURDAY
Proverbs 1:7 (King James Version)
7The fear of the LORD is the beginning of knowledge: but fools despise wisdom and instruction.
Proverbs 2:1-5 (King James Version)
1My son, if
thou wilt
receive My words, and
hide My commandments
with thee;
2So that
thou
incline thine ear unto wisdom, and
apply thine heart to understanding;
3Yea, if
thou
criest after knowledge, and
liftest up thy voice for understanding;
4If
thou
seekest her as silver, and
searchest for her as for hid treasures;
5Then
shalt
thou understand the fear of the LORD,
and
find the knowledge of God.
BIBLE STUDY AFTER LUNCH IN JUSTIN’S HOME
OPENING PRAYER TIMOTHY & DANIEL
LESSON Proverbs 1:7 & 2:1-5 MATTHEW
MEDITATION & DISCUSSION GROUP
A. Grammatically generally what is the relationship between the If clause and the Then clause?
B. Please give some examples using If/Then in our daily life in home.
C. Please give some examples using If/Then in our daily life at school.
D. Please give some examples using If/Then in our daily life in church.
E. Please give some examples using If/Then in our daily life at work.
F. What is the relationship between Prov. 1:7 and 2:1-5?
G. May we use Prov. 2:1-5 to help us understand Prov. 1:7?
H. In order to understand “the fear of the LORD” what will be involved?
I. Is your ear involved? How?
J. Is your heart involved? How?
K. Is your mouth involved? How?
L. Is your throat involved? How?
M. Is your hand involved? How?
N. Is your foot involved? How?
O. After you have done Items I through N, are you involved holistically (relating to wholes)?
P. Can we incline our ear & apply our heart by our own effort?
Q. Did you receive your salvation & eternal life from God free?
R. Will you achieve your spiritual maturity, sanctification, free?
S. Do you understand the “cleave to our LORD” mentioned in Deuteronomy 30:20?
T. Do you understand the “Ruth cleaved to her mother-in-law” in Ruth 1:14?
U. Did Ruth cleave to Naomi before she became her daughter-in-law?
V. Why did Ruth become to cleave to her mother-in-law?
W. How did Ruth and Naomi achieve that intimate relationship?
X. Do you know to build up an intimate relationship with your LORD Jesus Christ?
Y. What are the two basic actions we have to take in order to have an intimate relationship with our LORD Jesus Christ?
Z. Are you willing to practice all this in your daily life?
SUMMARY & APPLICATION JOSEPH & JUSTIN & MICHAEL
CLOSING PRAYER TIFFANY & JOYCE
Input provided by Matthew & Joseph
From: Matthew Huang [mailto:matthewyhuang@hotmail.com]
Sent: Friday, January 13, 2006 11:28 AM
To: Huang, Hsien L
Cc: matthewyhuang@gmail.com; matthewyhuang@hotmail.com
Subject: RE: 2006Jan14SaturdayMessageToGrandchildren.doc
Hi Grandpa,
Here it is.
Proverbs 1: 7
If you listen to God's words, you will have the true knowledge. The stupid people don't trust the
Lord and they don't have knowledge.
Proverbers 2: 1-5
If you listen to God and keep His words in your heart, you will understand what He said. If you
really want to know what God want's you to do, pray to Him. If you look for God's knowledge like searching for gold, then you shall understand what God want's you to do and the knowledge of God.
From: Joe boe [mailto:demurejoe@hotmail.com]
Sent: Friday, January 13, 2006 11:25 PM
To: Huang, Hsien L
Subject: RE: 2006Jan14SaturdayMessageToGrandchildren.doc
Hello Grandpa,
The teachings of these verses of Proverbs tells us to seek wisdom and heed
the teachings of those that are wise, but ultimately to fear the Lord. This
is because the Lord is the sole giver of all knowledge and to believe that
knowledge comes from another source is foolish, therefore first fear God and
heed to his teachings to be wise.
-Joseph Lin
Some of our fifteen grandchildren in Houston provided Answers to
Meditation & Discussion of our Bible Study
A. Grammatically generally what is the relationship between the If clause and the Then clause?
If clause gives the condition or requirement, Then clause gives the result. In other words, you must do such things stated in the If Clause before you can get such things stated in the Then Clause.
B through E. Several grandchildren provided different practical examples.
F. What is the relationship between Prov. 1:7 and 2:1-5?
Prov. 1:7 tells us the beginning of knowledge IS the fear of the LORD. Prov. 2:1-5 tells us what is the fear of the LORD and how to fear the LORD. They are closely related.
G. May we use Prov. 2:1-5 to help us understand Prov. 1:7?
Yes.
H. In order to understand “the fear of the LORD” what will be involved?
The answers to Items I through N answer this question.
Answers to I through N:
Elements involved How
Your ear Incline your ear
Your heart Apply your heart
Your mouth Criest – Pray and Praise
Your throat Lift up your voice – Pray & Praise
Your hand Use your hand to Seek and search
Your foot Walk on your foot to search & seek
O. After you have done Items I through N, are you involved holistically (relating to wholes)?
Yes. Your body, your soul, and your spirit must be all involved.
Never be absent-minded, nor be daydreaming, especially
when you are in the sanctuary and classroom. Don’t be just
providing only your physical presence there.
P. Can we incline our ear & apply our heart by our own effort?
No. John 3:16 says For God so loved the world
that He gave His only begotten Son
…. This tells us: He gave His only begotten Son to the world …
because He loved the world.
Prov. 2:1-2 says … thou so wilt receive God’s words &
hide God’s commandments
with thee
that thou incline thine ear…, and
apply thine heart to…
This tells us: you incline your ear unto wisdom, and
apply your heart to understanding
because you will receive God’s words &
hide God’s commandments
with you.
Q. Did you receive your salvation & eternal life from God free?
Yes. Eph. 2:8-9; Rom.8:32; Rom.5:8-9
R. Will you achieve your spiritual maturity, sanctification, free?
No. Phil.3:12-13; Heb. 6:1; Prov. 2:1-5
S. Do you understand the “cleave to our LORD” mentioned in Deuteronomy 30:20?
It seems abstract.
T. Do you understand the “Ruth cleaved to her mother-in-law”
in Ruth 1:14?
It is real. When you went to UT to study, the first time you left home, you felt homesick, you cleaved to your mom, your dad and your family members. You missed them.
U. Did Ruth cleave to Naomi before she became her daughter-in-law?
No. They were strangers to each other.
V. Why did Ruth become to cleave to her mother-in-law?
Because Ruth was married to Naomi’s son. They became family members to each other.
W. How did Ruth and Naomi achieve that intimate relationship?
They communicated more and fellowshipped more daily in their family.
X. Do you know how to build up an intimate relationship with your
LORD Jesus Christ?
Yes. Like Ruth and Naomi.
Y. What are the two basic actions we have to take in order to
have an intimate relationship with our LORD Jesus Christ?
Read God’s words, the Bible, and listen to the talk of our LORD Jesus Christ more often through the Bible, other person’s talk, or environment, etc, and talk to our LORD Jesus Christ more through prayer.
Z. Are you willing to practice all this in your daily life?
All grandchildren answered positively.
SUMMARY & APPLICATION JOSEPH & JUSTIN & MICHAEL DID VERY WELL.
CLOSING PRAYER TIFFANY & JOYCE DID VERY WELL
May God our LORD Jesus Christ bless you all.
The pictures below were taken by Joyce, your mom/aunt Chin, and your grandma. Most of the dishes, Chinese style and American style, were cooked by Justin. The dishes were delicious. Grandma gave Justin a monetary award. May all the honor, glory, praise, and thanksgiving be to our LORD Jesus Christ God only. In Christ name, AMEN.
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