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Here is a description of the principle of the operation of a new lightning device:

During a storm the ambient electric field may rise to between 10 to 20 kV/m. As soon as the field exceeds a thresold representing the minimum risk of a lightning strike, the OPR lightning terminal is activated. It draw its energy from the ambient electric field the energy required to generate high voltage pulses, creating and propagating an upward leader. No other power sources are required, and no radioactive components are used.

It remains unclear to me the physics of the drawing the energy from the electric field. Obviously, you can not draw the energy from the electrostatic field, no matter how strong it is. So i suppose that the electric field must oscillate more or less. The question is:

Which is a minimal frequency of the oscillations so that such device can work properly?

A full description of the device

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Dear Martin, if you think that you cannot draw energy from an electrostatic field, how do you explain that some people claim that they use batteries in their cell phones or even laptops? ;-) Is that unimaginable that the electrostatic fields allow one to lift charged objects from one place to another, doing work along the way, and reducing the strength of the electrostatic field (which is why batteries get discharged)? –  Luboš Motl May 11 '11 at 8:28
    
Lubos: While I am agreement that one can draw energy from an elctrostatic field by moving charge about, the change in charge distribution reduces the loval field strength. It seems to me there isn't much energy available given the quoted field strengths, and a reasonable guess as to divice size. Perhaps Martin has a point, it seems unlikely to me that such a device could draw enough energy to work. And I think pre-strike charge buildup is slow and one-way, so you don't have a high frequency to work with. The electrostatic energy in the battery is very small, chemical energy dominates. –  Omega Centauri May 11 '11 at 16:58
    
@Omega Centauri note the quote above, '10 to 20 kV/m' during a storm. Some numbers here: en.wikipedia.org/wiki/Paschen%27s_law . For small gaps the breakdown needs smaller voltages. I see no conceptual difficulty in the setup using the kilovolt potential difference as a battery to generate the pulses it needs to break down the air and create a leader. –  anna v May 11 '11 at 18:42
    
@Luboš: A battery is not a good example. In this case the integral of the electric field around a closed loop is nonzero: $\varepsilon =\oint \overrightarrow{E}\cdot d\overrightarrow{l}$ (electromotive force of the battery). This quantity is zero for electrostatic field. –  Martin Gales May 12 '11 at 6:07
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2 Answers

The quotes you give do not really explain much.

With a conductor one can short the static field of the atmosphere, one does not need oscillations for that. I think it was Benjamin Franklin with reaching the clouds who discovered that first hand?

Lightening strikes when naturally induced upward leaders reach and meet the charged cloud downward leaders creating a path of ions for the energy to go through. This happens when the fields are strong to strip neutral molecules from their electrons and create an ionisation path.

Edit, correction: Just saw the link you provided. It seems that it works in channelling the lightning to the rod instead of playing russian roulette on where it will hit. It is the logic of simple lightning rods, so that the energy is sent to the ground without damaging the buildings. The proposed design makes sure that this will happen instead of leaving it to the probable fields that would develop naturally with the simple lightning rods.

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You can draw energy from electrostatic field only if you apply an external(not due to the field) force, i think. By description, there is no such force for the device. –  Martin Gales May 12 '11 at 6:09
    
@Martin Gales Why? It is like a DC battery: ground - atmosphere potential difference will be applied on the + (1meter higher) for example, and - input of the circuits designed to create the pulses.The energy will be continuously supplied by the storm. –  anna v May 13 '11 at 7:21
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Whether the mentioned device is effective or not I can not guarantee (Although I don't see any problem in the working of the device and its basic principle seems to be more or less same as an ordinary lightning arrestor).

Let me slightly elaborate the way a lightning arrestor works in general.

A lightning arrestor generally consists of a metallic rod with a sharp pointed terminal, called prong, at the upper end and it is fixed at the highest point of a building. It is grounded by a good conductor.

Whenever a charged cloud comes above the device it induces opposite charges in the upper portion of the lightning arrestor. Since the upper terminal is pointed the charge density is maximum there and a lot of charge leakage takes place from the prong.

Those charges move upward by Coulomb attraction and neutralize the cloud as much as possible so that the chance of a lightning strike becomes less. However for a heavily charged cloud the discharge phenomenon is so fast that before it is neutralized fully by the upward moving leakage charges of opposite sign, a lightning strike takes place.

Since the L.A. is installed at a high enough place, the chances of catching a lightning strike by it is high, since it provides a path of very low resistance to the ground. In other words, it provides a safe passage to the very high electrical discharge thereby saving the building. Normally it safeguards an area which lies within an imaginary inverted cone where the tip of the cone is on the point of the prong. Apology for the bad drawing by me on paint brush :)

enter image description here

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I googeled for that "ESE" devices yesterday, to find lab tasts and practical tests on high buildings in Kuala Lumpur. All resulted in: no significant difference to a traditional "Franklin rod". –  Georg May 12 '11 at 8:39
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