Cockroft-Walton question.

don't know about scaling but i do know it makes nice plasma, the ozone is nice especially when coming in contact with moisture! generates some nice nitric acid that will make your skin do alittle etching .!

Suppose you had a light, high-voltage, high-current power supply. You connect one power supply terminal to the lifter body. Where do you connect the other?

John

This page shows construction of a lifter:

How to build an HexaLifter for your experiments.

One lead from the power supply is connected to wires at the top of the lifter. The other lead is connected to metal foil at the bottom. You have insulators separating the wires from the foil. Then the air between the wires and foil serves as a dielectric for a capacitor. The asymmetric geometry of the two sides of the capacitor, the wires compared to the flat metal foil, causes a flow of ions from one to the other.

Bob Clark

This may be a wild idea, but how about finding the smallest, lightest rechargable cell that has the power/weight ratio you need, making a big stack of them, and from the fairly high voltage you get driving a single DC-AC inverter followed by a single Cockroft-Walton? Adjust the size of the battery stack and the size of the Cockroft-Walton to give you the best power/weight ratio. I havent run the numbers, but it might turn out to be good enough to fly.

If my CW variant that directs the output of one CW into the input of another is to work, I need a way of converting high voltage DC to AC. The 555 oscillator IC used in a DC to AC inverter is not designed for high voltages. Some possibilities are suggested by the graphs on this page:

Experiment 15 - Cockroft Walton Multiplier.

Firstly, does the input voltage have to have a negative portion? Look at the image near the middle of the page showing a graph of the outputs from each of the capacitors in a 3-stage CW. Note that even numbered capacitors have a sine voltage, it's just shifted upwards to always be positive. Nevertheless this sine wave gets sent to the next capacitor in this CW to result in a more or less flat DC voltage. Perhaps a CW can therefore take a varying sine wave voltage input even if it's never negative.

Secondly, take a look at the image near the bottom showing the ripple voltage in a CW output. The ripple is large like this when the frequency or capacitance is low. However you can make the ripple be small by choosing high rated capacitors. So one idea is to have two CW's at each level, one that has small capacitors to get large ripple and one with large capacitors to get small ripple. You want this second CW with low ripple to have output voltage right in the middle between the peaks of the CW with the large ripple. Then you just take the voltage difference between these two CW's to get an AC voltage: half the time the voltage is positive when the 1st CW voltage is higher, and half the time it's negative when the 1st CW voltage is lower. But you need the 2nd CW voltage to be right in the middle of the first. Normally it should be *higher* than the upper peak of the 1st CW with the high ripple. Perhaps you could use a diode with a high voltage drop to bring this 2nd CW down to the middle between the peaks of the 1st CW. The disadvantage of this method is that it requires two CW's at each level. Is there some simpler method to drop down a varying sine wave voltage to make the center be at 0 volts?

Bob Clark

But that's not the way the hexalifter is wired. External (grounded)

If you did make an unthethered lifter, with no return current path to ground, its ion emission would quickly build up a huge net potential on the whole structure. Has anybody ever got this to work?

There's a much more efficient way to convert electrical energy to lift; a helicopter rotor.

John

I'd think the most challenging part of that would be finding the AC batteries. ;-P

Perhaps the second wire connection is not visible on that page, but all lifters are made in this same way. See for example this page:

How to build a Lifter v2.0 demonstrator. "The Lifter v2.0 is finished, the +30kV is connected to the wires through a thin copper wire ( 1/10 mm ) on its left side as show on the photo above. On the opposite side, the aluminum plate is connected to the 0V ( or the ground ) with a small aluminum tape. Your HV power supply must be able to deliver about 500 =B5A at about 30KV."

See also the many examples worldwide on this page:

The Worldwide Lifters replications Log Book.

Lifters are not nearly as energy efficient as propeller driven craft

Bob Clark

I'm working on learning to fart with 175 lb of thrust. ;-P

I am connected to an east coast university. But my background is in pure math not electrical engineering. I wanted to get some feedback on the feasibility of this idea first. Not many faculty around over the Summer, but I will see what they say over there.

Bob Clark

Here are a couple of comments:

Once you have decided to use a DC-AC inverter, you are no longer talking about sine waves. It takes extra effort to generate sine waves and the don't work as well as square waves in this application.

I advise dropping your idea about using the ripple as a way to get AC from a DC-output Cockroft-Walton so as to drive another one. That idea is a dead end. If your power supply is a battery you need a DC-AC inverter. There is no way to get around it.

Bob, I think at this point that you need an electronics engineer to sit down with you and help you with what you are trying to do. I don't think that newsroup posts are enough to get you where you want to go. Is there a school nearby with an electrical engineering department? What part of the world are you posting from?

LOL! Well, there is the reaction mass issue. ;-P

I used to have a terrible problem with Newark. I would get orders with each line item in a seperate shipment from the *same* factory. The shipping charges were sometimes more than the cost of the components. I finally got my sales contact to put special 'ship in one box' instructions on all my PO's.

Bob

no s*it?

Cheers Terry

The hexlifter page says that one wire supplies hv to the lifter's upper wire array, and that it "is maintained with three thin nylon wires to the styrofoam base plate" which doesn't sound conductive to me.

So the single uplink wire replentishes the charge lost by creating ions. If you float the power supply onboard, and still eject ions, a huge charge will quickly build up on the structure.

Has anybody ever built a self-powered, unthethered lifter?

John

Probably not. A lifter needs to iozize molecules. This takes a fixed amount of energy per molecule.

Only then can the molecules be expelled. Expelling at high speed means you need few ions but waste a lot of energy as kinetic energy of the ions.

Expelling at low speed means ionizing a lot of molecules because each will give only so much impulse.

If you have a lot of energy and not a lot of mass, ions are a good propulsion method: see what Toshiba and Mitubishi do (google "ion engine").

If a mix of ions and air is expelled things may work better. However this is not a matter of the power supply...

Thomas

The spec sheet for the DC-AC inverter specifies *DC* battereies..

My point exactly. An untethered, self-powered lifter would have charge problems.

John

One reached moon orbit from a transfer orbit. It took 18 months... and wasn't a real lifter because it used xenon that it carried along and not atmospheric air.

The mass-to-thrust rations of there are lousy: often tenths of a newton, leading to milli-G acceleration.

This is the closest I suppose.

BTW the ion engines use an electron gun to stay electrically neutral - or all the ios wouldk be attracted back into the vessel, damaging it and zeroing the performance of the engine.

Thomas