How do ionic wind Lifters really work?

  1. So I've read about these Lifters that lots of crazies attribute to antigravity or whatever. Obviously we know that it's caused by the ionic wind.

    But how does that work, exactly?

    One description I read says that the high voltage of the top conductor ionizes air (by removing or adding electrons?), which is then attracted to the bottom conductor. On its way to the bottom conductor it collides with neutral air molecules, and imparts momentum to them. The neutral molecules are not attracted to the conductor, so they just keep on moving in the same direction, creating the wind. The charged particles keep going until they hit the bottom electrode, give up their charge, and become free-floating neutral particles again.

    (If this is the correct way to think about it, how is power transferred from the source to the load? What effect does the change in momentum of the moving particles have on the electrical properties of the air? Do the collisions look like added resistance to the circuit? Could you stop the air from moving and see a change in resistance?)

    They say that the lifters don't work as well if they're arcing, which would seem to mean that the corona discharge works better because the charged particles are more dispersed and interact with neutral air molecules more often, so this description makes sense to me.

    Another description seems to say that the force seems to come from neutral air molecules becoming polarized and attracted to the top conductor, then being repelled from the bottom conductor. Is this really just the same thing as above? It almost looks like it's supposed to be expelling a constant stream of negatively charged ions, which can't possibly be true, can it? The lifter and power supply would then become more and more positively charged, and the negative particles would flow right back.

    Things like this and those "ionic breeze" air "purifiers" claim that they release ions into the general atmosphere surrounding the unit. But is that really true? Wouldn't the ions just travel from one electrode to the other? Maybe they really mean the byproducts of ionization like ozone...

    Also, stop calling them "asymmetrical capacitors". Capacitors have this thing called an insulator between the two plates that prevents charge from flowing. Ionized air is not an insulator.

    I saw this in a related thread from 2004:

    That's a really awful attitude. Don't censor it; debunk it. Otherwise they'll never learn and keep coming back.

    Haven't you ever been to a crackpot website? "Go to our website to find out what They don't want you to know!!!" Censoring them just makes them even more crazy.
  2. jcsd
  3. Why? They ARE asymmetrical.
  4. I never said they weren't.
  5. The ultimate test of the "Lifter" would be in a vacuum.
  6. Yeah. They've been tested in vacuum and they don't work.

    As I showed in one of my links...
  7. I would suppose,then, that the "Lifter" device primarily operates under ionic wind dynamics.
  8. Yep. As I said.

    So how do ionic wind dynamics work?
  9. It seems like your first description in your original post is essentially correct, though I am no expert.
    What I do know is that the top wire of the "Lifter" ionizes the surrounding gasous environment, and that the bottom, oppositely charged "electrode" section of the Lifter is electrically attracted towards that upper ionized environment, resulting in movement of both the lifter and the ionized gas.
  10. pervect

    pervect 8,156
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    There's a simpler somewhat different version of dipole explanation that I think is likely to explain the lifter. Ionization is not necesarily involved.

    If you have a dipole


    the force on the dipole is proportioanl to the rate of change of the electric field. This is because the total force on the dipole is E(right)*q - E(left)*q. If E(right), the electric field on the right side of the dipole = E(left), there is no force. This is true for a uniform electric field. If the field is not uniform, however, there is a force on the dipole, which is proportional to the rate of change of the field strength.

    The assymetrical capacitor will generate an electrical field with a gradient. (It will be stronger near the small plate, and weaker near the large plate).

    This is not true of a symmetrical capacitor, which will generate a uniform field between the plates.

    So we would expect an induced dipole moment on the air, and the induced dipoles should follow the electric field gradient- but only with an assymetrical capacitor.

    To test whether this proposed explanation actually is correct would need some experimental verification.
  11. Ivan Seeking

    Ivan Seeking 12,122
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    That's an interesting point, pervect.
  12. I don't think that's correct. That strikes me as perpetual motion machine thinking. Hmm...

    Doesn't seem right to me, either.

    It won't be any stronger, it will just be "less dense". The field lines would be spaced farther apart in some regions and closer together in others. I wish I remembered my electromagnetics... I'll dig out my books if I have to. :-)

    Highly skeptical. :-) I don't think there really is such a thing as an "asymmetrical capacitor". Think about it this way. Put two metal plates very close to each other in a typical capacitor situation. Now imagine a variable capacitor, in which the plates slide so that less of them is facing each other. The "overlap area" becomes less for both plates. Now imagine the same thing with one smaller plate on top of one bigger plate. The area directly across from the smaller plate acts as a capacitor, while the extra non-overlap isn't really part of the capacitor. I know this is an oversimplification and hard to explain, but I don't think the "asymmetrical capacitor" is right at all.

    What would you propose for a test? Immerse it in a material that has all qualities (viscosity, etc.) equal, but with differing tendencies of the molecules to form dipoles? Testing with and without corona discharge?
  13. Posters here seem to operate as though this subject exists in a vacuum. Ionic wind is a well known phenomenon that is currently exploited in an ever increasing range of products and with ever improving efficiency. View, for example, the 8 patents of Krichtafovitch assigned to Kronos Air Technologies, which uses them to move up to 20,000 CFM at speeds up to 1500 FPS.

    Regarding the lifter in an "hermetically sealed" environment, hermetic sealing refers to movement of solids, fluids and gasses. It does not affect, for example, radiations, magnetism, or in this case capacitance. In fact, the plastic film making the hermetic package is a well known dialectric useful for capacitance. Imagine charges at opposite ends of that dialectric having charges in the tens of thousands of volts potential. The air outside that film will be electrically charged with potentials having somewhat lower values. Those potentials will be fully capable of generating an ionic wind effect on the outside of the system.

    To properly test this hermetic seal it would have to be surrounded by an electric shield of very low resistance that also had a near zero resistance and inductance to ground. Such a test hs not been done.
    Last edited: Jul 30, 2005
  14. I've been looking for something like that for years! Hooray that someone finally did it.

    I've been learning a lot more about corona lately, and the corona particles impinging on neutral air molecules makes a lot more sense now. I'm still not sure:

    • What happens when the corona is local. In other words, corona forms anywhere the electric field strength exceeds the breakdown of air, which is about 30 kV/cm. With certain geometries, the conductive corona forms only directly around the conductor, and, although it effectively increases the size of the conductor, the electric field strength around it is not high enough for the corona to expand any further. So the conductive region sticks around the conductor and there is no spark-over. So how is current transferred through the still-insulating air in between two conductors with local coronas? Or is there no more conduction than normal? The corona creates ions which stay separated as ions and then diffuse through the neutral air gap? I guess that's what happens...
    • What the high-voltage source "sees" when a corona has formed. Does it just appear as a resistance? Do the particles have momentum which appears as inductance?
  15. There was an academic paper about lifters in a recent issue of the Journal of Electrostatics. No mystery, but I was a little surprised at some of the references. Academics should ignore 'internet crackpots'.
  16. pervect

    pervect 8,156
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    I've seriously neglected this thread. Anyway, to clarify something (in case anyone is still listening) - to experimentially distinguish between the ion hypothesis and the induced dipole hypothesis that I mentioned, it would be necessary to do an experiment to see if the exhaust of the lifter actually contained *ions*. I'm not aware of anyone doing this, even the previous poster who built one.

    This could be done by seeing if a gold-leaf electroscope was discharged by the exhuast, for instance.

    If the exhaust did contain ions under normal room conditions, it would still be interesting to see if it still contained ions when care was taken to use very pure, dry air (no dust particles or water droplets) - I.e. it's possible that the exhaust might contain some ions normally, but that they aren't critical for generating lift.

    For those posters needing a reference on the fact that there is no force on a dipole in a uniform electric field, but that there *is* a force on a dipole in a non-uniform electric field, I'll give the following reference:

    On the topic of induced dipole moments,

    Air has a relative permittivity of 1.0006 according to google, so the effect is small.
  17. For the first part, I think the electrons are pulled off the top wire (by the bottom plate), and repelled away from the wire with a force F=2kqa/r, where a=charge per unit length (because it's a line charge, although the equation is ideally an infinate line), and then attracted to the vertical lower plate by F=2k(pi)qb, where b=charge per unit area (because it's a charge on a plate, although the equation is ideally an infinate sheet). So the plate has a force pulling the electrons from the wire with less dependancy on the distance from the moving electron than the wire does. Another way to look at it is, when an electron is thrown from the wire in a near verticaly upward direction, it feels repultion from the line charge in a more localized area than the force from the plate, so it will get pulled around the wire and attracted to the plate, and get repelled from the wire at the same time, so it has a chance of zooming past the plate, or at least bumpung into some neutral air molecules and send them flying off in a downward direction and then get sucked into the plate. The main thing to consider is that the electric field around the whole apperatus is such that it shoots electrons out into the air, and then sucks them around to the bottom, and with so many electrons transferring thier momentum to air molecules in an assymetric way, there emerges a net force vector that pushes the apparatus through the air, or should I say, pushes the air through the apparatus.

    I think I over complicated things, and Pervect explained the main thing I am trying to say, that "that there *is* a force on a dipole in a non-uniform electric field, I'll give the following reference:" - Pervect
    Last edited: Oct 27, 2005
  18. DaveC426913

    DaveC426913 16,537
    Gold Member

    The only thing that will accomplish is to alienate the rest of the folk. A forum devoted to rational and critical discourse will be diluted and polluted by crack pottery.

    Doubly so on the Web: you can't change it - all you can do is protect your corner and ignore any intrusions.
  19. Ordinarily, Electrons would need a target with charge difference to seek an equalibrium and make all charge differences neutral.

    But, If the emitter sends a charge potential of say +50keV to a (potential) target and the (Target) is switched off before the Electron charge reaches the other side, the Electrons should become mean free Electrons.

    Lifters in a vacuum would have to utilize such or other techniques to convert the Electrons to mean free particles.

    There are other probabilties like specific angle trajectories of the Electrons striking the (Aft) of the target rather than the (Bow) of the target allowing the Electrons to produce a Net (Push) with a mass of 1 per Electron against the rear of a target.:smile:
  20. pervect

    pervect 8,156
    Staff Emeritus
    Science Advisor

    If the "lifter" generates electrons, those electrons will also discharge an electroscope.

    Hoever, it is rather unlikely that electrons cause the lift, because they have little mass. If electrons are generated, positive ions will also be generated. Having much more mass, they will have a much higher momentum for a given velocity, or for a given energy.

    I've also attempted to explain why the generation of ions or electrons *may* not be necessary to generate lift.

    I wouldn't rule out ions as the source of lift, by any means, but at this point I see two plausible (and purely electromagnetic) explanations for how the device could work, one of which involves ions, and one of which involves only dipoles.
    Last edited: Oct 29, 2005
  21. glad than we can talk with a lifter user.
    Here a variation based on Townsen T Brown original work
    You can even try under vaccuum
    Hint use a tension variable 50 KV to 250 KV
    (a) Observe varying velocity with varying applied voltages (50 KV-250 KV dc)
    (b) Observe varying effects of ion emission and corona discharge colorations at varying vacuum chamber pressures.
    Copy of a letter from Thomas Thownsed Brown about tests made under vacuum chamber.

    Edited to add
    Biefield Brown effect experiment

    Other in english

    Last edited: Nov 2, 2005
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