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11 May 2024 19:17
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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: |
James |
| Subject: |
Anyone know how to build a propulsion device? |
| Question: |
Does anyone really know how to build a propulsion device which can hover in air? I've heard people said that it can be done by spinning the solenoids in certain way. Any comments? |
| Date: |
20 January 2004
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Answers (Ordered by Date)
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| Answer: |
fred groth - 20/01/2004 06:43:45
| | | hello there!
i built a rather crude prototype based on a trapped sphere
traveling in an elliptical orbit. i took an aluminum block and cut an egg shaped
channel in it. Then i placed a 1" diameter ball in the channel and covered it with a plexiglass sheet. i then put two holes in the side of the block connected to the channel,
one for air input,one for exhaust. i then applied 130 psi (all i had) to the input port.
the results were interesting. it kind of wiggled from point A to point B, but it did traverse
as expected. like i said it was crude but there are many ways i can
think of to stabilize and increase power. next time i will turn it skyward
and take out some lights in my shop!
if only i can find the time.....
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| Answer: |
Brad H - 20/01/2004 20:36:54
| | | No one has produced any solid evidence that rotational motion can be converted to linear motion without expelling matter (newton's third law). As a matter of fact, any purely mechanical system that claims to achieve this feat is either violating physics principals, using some unknown physics laws, or is simply not working. The reason that little real scientific research is being done in this area is because most people in the the field believe it is not possible.
A reasonable alternative to gyroscopic thrust for space travel is producing super-efficient electrical rockets. Sometimes called 'ionic propulsion' these rockets use electricity and magnetic principals to accelerate particles to fantastic speeds (like the Ionic Breeze air filter, only more powerful). The faster the particles are expelled, the more thrust is achieved. In theory, these particles could be accelerated closer and closer to the speed of light, thus providing more and more thrust per unit of matter. Eventually the process could be so efficient that fuel tanks for extreemly long trips could be very small indeed. Imagine getting 1,000,000,000 km per litre of fuel or more. Though that's a radical number, current physics does NOT exclude this, it supports it.
That's not to say that research in gyroscopic/inertial thrust is wasted. It could very well be that we will discover a method for creating an unbalanced force in a system without outside interaction, but I believe such a breakthrough will be the result of theoretical advancements, and not experimentational research. If you recall, the Atom Bomb was a direct result from discovered physics principals, not from chance experimentation. Only time will show who's correct!
Keep on inventing, and good luck!
Brad H
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Nitro MacMad - 24/01/2004 09:39:55
| | | Dear James,
Your question needs answering on two parts. Yes, I really know how to build a propulsion device. No, it cannot hover in the air. Could you explain what you mean by spinning the solenoids as I am not sure on that.
The reason that hovering in the air is not possible (at least at the moment) is that the device is gyro based and *reactionlessly converts torque into a linear displacement of its c of g. To maximise the amount of precession, which is what is used to carry out the "impossible conversion", you have to maximise the giros mass, maximise the gyros diameter, and maximise the gyros spin speed (sorry, Dr. Fisher, angular momemtum). All of this conspires to make the machine very heavy in relation to its "thrust" (I am tempted to call it "Impulse" as that really is a better discription of what it does but it would sound like science imitating space opera) and thus make hovering in the air rather unlikely.
However one should never say never (so I am told). You can exchange mass for spin speed in gyros and some years ago carbon fibre brushes (rather like sweeps brushes) were spun at increadible speeds in vacuum housing to remove air drag. They were intended as energy stores but they would make wonderfully light precessing gyros so....
Kind regards
nm
*(that horrible word "reactionlessly" is actually wrong as the machine produces plenty of reactions and actions but importantly it ends up with "shed loads" of equal but only a "Gnats" of opposite)
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Dean Hunt - 27/03/2004 04:31:31
| | | Propulsion is possable using gyroscopes, Sandy Kidds machine is a good example of this, but there is a fundermental component missing from the design, Sandy said him self
that his machine only worked when he introdused a wobble on the bottom of the arms that link the gyros to the lower main shaft. By breaking that connection and fitting a pulling divice such as soliniods then you can pull the force of precession from the giros onto the lower shaft of the device. I did this test myself a device wieghing 1 stone pulled a force of three stone. However my engineering skills are not that of Sandys and it tends to be difficult to keep the device working for longer than 10 seconds. in my design
I used cams to pull down the precession but they just kept on snagging.
Another thing to take in to consideration is that you must let the gyros recover so if you choose to use neumatics or solinoids then you must pulse the pulling action has if you was using cams, try it it works
Regards
Dean
PS. did the dean machine work or not?
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| Answer: |
Eric James ----- - 29/04/2005 06:44:41
| | | James,
Yes. A helicopter is a gyroscopic device that hovers in the air. :)
Okay, I know that's not what you meant.
With gyros, I doubt what you are seeking is possible. There is (at best) only inconclusive evidence for even the merest propulsive effect. As far as I know, none of these inconclusive results exceeded the margins of error for the experiments in question.
Eric
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| Answer: |
Sandy Kidd - 12/05/2005 12:19:07
| | | Eric,
Like many others before you.
Did you ever go out of your way to find out
Sandy
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Vincent Jones - 06/08/2006 20:01:24
| | | James,
In my opinion, it is possible to build such a device using a solenoid. I can explain the principle and list the components. The rest is up to you.
ELECTRO-MECHANICAL NWEEL CYCLE
1. A mass is constrained to move around an axis (the axle is centered on the axis of rotation), such that it can be forced to move outward against spring pressure by a solenoid that is integral with the mass (the mass is mainly comprised of a solenoid coil and weighted sleave); which slows the angular velocity of the rotational assembly (including the mass), due to conservation of angular momentum.* The mass is forced outward against spring pressure when the peripheral contacts are closed, and the solenoid activates. 2. The mass is then constrained to move in a widened arch for 180-degrees, which is the aft part of the cycle. During this part of the cycle, a motor provides rotational impetus via a spragg or roller-ramp clutch; which is to make up for frictional losses. 3. When the mass has moved approximately 180-degrees, the peripheral contacts are opened and current through the solenoid is abruptly stopped; thus the spring is free to push the mass back toward the axis. 4. At this point the mass becomes constrained to move around in a smaller arch for 180-degrees, which is the forward part of the cycle. During this part of the cycle, friction is minimized (e.g. no contact is made with the casing, except through bearings); and the rotational assembly is expected to turn faster than the motor is turning the clutch. 5. The cycle begins again, with the peripheral contacts closing; such that the mass is once more forced outward by the solenoid. (To better understand the latter explanation, see a photo at the following site: http://www.geocities.com/impulsedrv/002.jpg Note that the solenoid is not activated in this photo, so that the spring would be more easily visible.)
* This is the same as an ice-skater slowing down after a pirhouette.
MATHEMATICAL EXPLANATION
From the equations for centripetal force and conservation of momentum; I derived an equation to compare the change in centripetal force with the change in angular velocity, which occur during the cycle. (These are idealized equations, because they are not being used to take into account the parts of the rotational assembly that do not move radially. Thus the actual change in angular velocity due to a change in centripetal force will be less than it would be for an independent mass.) The equation is as follows:
ω2 = ω1(Fc2 / Fc1) ^ 2/3
Since it is probably hard to read it on this website, I will describe and explain it. On the left side of the equation is Omega Two, which is the resultant angular velocity for a mass that has been subjected to a change in centripetal force. On the right side of the equation is Omega One times the quantity [(Centripetal Force Two / Centripetal Force One) ^ 2/3]. Omega One is the angular velocity of a mass, before it is subject to a change in centripetal force. An increase in centripetal force results in both a reduction of the orbital radius and an increase in angular velocity. If angular velocity were to change in direct proportion with a change in centripetal force, it would be expected that the amount of time for centripetal force to act would be decreased proportionally to the change in centripetal force; thus there would be no net impulse from a completed cycle (360-degrees). Yet, from the above equation, we can see that angular velocity does not change in direct proportion with a change in centripetal force. To demonstrate this more clearly, I rearrange the equation as follows:
ω2 / ω1 = (Fc2 / Fc1) ^ 2/3
Note carefully the fact the ratio of Omega Two to Omega One varies as the 2/3-power of the ratio of Centripetal Force Two to Centripetal Force One. This indicates very strongly to me that the nweel is based upon a valid principle; such that (for example) if Centripetal Force Two is double Centripetal Force One it follows that Omega Two is 1.587 times Omega One. Since a doubled force is acting for a period of time that is more than half its original period (1/1.587 = 0.630 as opposed to 1/2 = .50) a net impulse results.
You have a method and why it works. Now go on and try it out, if you so decide.
Cheers,
Vince
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| Answer: |
Jerry Volland - 11/08/2006 16:09:22
| | | I once built a device which had a circlular track whoose center was off set from the shaft. The shaft had a spoke which spun a weight with a hole through its center, and this allowed the weight to slide back and fourth along the spoke, compressing a spring which held the weight against the cam track.
Experimenting with this device revealed a number of usable effects. Perhaps the most notable was that when the weight pushed against the part of the cam track which had a decreasing radius, relative to the shaft, the increasing torque - at a very slow speed - caused the perpendicular motor to twist so that it pushed the track against the weight. This twisting produced a strong movement of the device, perpendicular to the motor's twist.
Analysing the effect indicates that such an ecentric cam, when subjected to an *external* force (such as horizontal movement), will produce a thrust which is perpendicular to the applied force. So a jet or airplane using this system should produce lift without wings, as long as the external, horizontal speed is maintained.
JV
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Arthur Dent - 17/08/2006 10:04:15
| | | If one takes solenoid-shaped objects, spins them about their own axes and then connects several radially to a vertical shaft and rotates that about its own axis, one will get lift-off. However, this is not levitation. It is called a Magnus-effect helicopter, and works only in an atmosphere. The same principle was used to propel ships back in the 1920s.
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Jerry Volland - 17/08/2006 17:48:39
| | | You can also get lift by spinning a horizontal rod which is traveling horizontally, perpendicular to its length. Likewise, a number of spinning rods radiating from the edge of a disc will produce lift when the disc spins.
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Jerry Volland - 17/08/2006 17:51:42
| | | I wonder if James knows that his thread is active again. Since he didn't include his email, does anyone know how to contact him?
jv@spaceoffice.us
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| Answer: |
Vincent Jones - 29/08/2006 20:02:01
| | | NWEEL MATH
Here is how I came up with the nweel equation:
Since angular momentum remains constant when a centripetal force changes on a mass moving in an orbital path, the foregoing equality is known to be true:
m (r2 ^ 2) (w2) = m (r1 ^ 2) (w1)
(conservation of angular momentum)
…where m = mass, r = radius, and w = angular velocity.
The mass cancels (because it is equal on both sides of the equation), leaving…
(r2 ^ 2) (w2) = (r1 ^ 2) (w1)
Furthermore, we already know that centripetal force can be found as follows:
Fcn = m (rn) (wn ^ 2)
…where Fcn = centripetal force. Thus we can solve for r, giving…
rn = Fcn / m (wn ^ 2)
Substituting into the former equality, we arrive at:
[Fc2 / m (w2 ^ 2)] ^ 2 (w2) = [Fc1 / m (w1 ^ 2)] ^ 2 (w1)
and the process of solving for w2 (mass cancels) gives us…
w2 = w1 (Fc2 / Fc1) ^ 2/3
or...
w2 / w1 = (Fc2 / Fc1) ^ 2/3
Substituting numbers into the equation w2 / w1 = (Fc2 / Fc1) ^ 2/3 we find out that a change in angular velocity does not correspond one-to-one as to the implemental change in centripetal force. For example, if the centripetal force is doubled; then the angular velocity increases only about sixty percent. Thus we find that there is an indication that there may be a tendency for greater impulse to be imparted by the forward centripetal force of the aforementioned system than the aft force.
THIS IS IN REGARD TO A GYROSCOPIC PROPULSION DEVICE THAT UTILIZES THE GYRATION COMBINED WITH A RECIPROCATION OF A MOVABLE MASS (SUCH AS A SOLENOID) TO PRODUCE A PROPULSIVE FORCE.
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Vince Jones - 02/09/2006 22:56:45
| | | CALCULATION OF THE NET FORCE PRODUCED BY A NWEEL
In order to calculate the [approximate] net force produced by a nweel, we can use the following equation:
Fnet = Fc2 x Tr x wavg x THE INTEGRAL OF SINE FROM ZERO TO PI
Now, I don't know about you all; but I come up with 2 for the integral of sine from zero to pi. Like I said, it will come out to within an order of magnitude; even if I am mistaken somewhat. Simplifying the above equation, thereby we come up with:
Fnet = 2 x Fc2 x Tr x wavg
...where wavg is the average of w1 and w2.
It is a little over my head to explain how I came up with this equation, but I will try my best: Fc2 (centripetal force two) produces an impulse via being multiplied by Tr (residual period) which is a vector sum with a duration of 180-degrees (i.e. pi radians). [The latter is where the integration of sine comes in.] Furthermore, this impulse occurs once per every revolution---and each revolution occurs as an average of w1 and w2---so the units of rev/sec and sec/rev conveniently cancel each other out; and we are left with the units of good old kilogram-meters-per-second-per-second. [BTW, there should definitely be a better equation for the net force of a nweel (i.e. it would ideally take into account all mass that comes into play within the rotational assembly), but I have not come up with it yet.]
N.B. The calculation of net force produced by a nweel involves some basic calculus, so I may have mistakes here; whereas it is fairly inconsequential, since the numbers produced with the equation will still fall within an order of magnitude of actual results.
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Arthur Dent - 05/09/2006 14:08:55
| | | Dear Mr Jones,
somebody has already invented an energy-producing device which is based upon similar calculations to your own. Since this energy 'comes from nowhere', his calculations are obviously wrong.
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Arthur Dent - 05/09/2006 14:09:33
| | | Dear Mr Jones,
somebody has already invented an energy-producing device which is based upon similar calculations to your own. Since this energy 'comes from nowhere', his calculations are obviously wrong.
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