Cockroft-Walton question.

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

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?

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

no s*it?

Cheers Terry

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

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

Yup. The old payload vs. propellant quandry.

Bob

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

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

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

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

John

John, I really don't see why you're arging this point. The "thin nylon wires" are used to hold the lifter in place so it won't fly away. They are not the conducting wires for the power supply. Lots of amateurs have made lifters with no tethers: they usually fly up far enough to disconnect the power supply wires or damage the lifter. There are many web sites online discussing the lifters which you can find by Googling. You can write the authors of these web pages if you don't believe how they work.

Bob Clark

Because I'm an engineer.

But can they fly with *no* wires?

John

Because he is under the mistaken impression that he is an engineer. A real engineer has the all-impotant attribute of listening to what people tell him. You can tell Larkin how lifters work and explain that the only reason they have wires going to them is because the power supply is too heavy to lift all day, but it won't get you anything but questions and - eventually - personal abuse. Ignore him.

Have you calculated whether a big stack of coin cells driving a DC-AC inverter and then a Cockroft-Walton stack will be light enough to fly yet? what number are you getting?

You are wasting your time explaining. Usenet being what it is, I strongly advise you to look at how helpful each person who responds to you is and to not reply to the non-helpful. I even wrote up a nice explanation of how this works so folks can quote it. Here it is:

THE STANDARD ADVICE:

There is a way to influence what gets discussed in a newsgroup that works well, and another way that has never worked no matter how many people have tried it.

What works: Post articles on the topic you wish to see discussed, and participate in the resulting discussion. Use killfiles and filters so that you don't see the articles that you dislike. If you don't know how to use a killfile, use good old fashioned discipline and don't read the articles that you dislike. Never, ever respond to articles that you dislike.

What doesn't work: Respond to articles that you dislike, complain about articles that you dislike, complain about posters that you dislike, complain about how terrible everyone else is for not posting what you want them to post. Talk about how to respond to articles that you dislike. Make the articles that you dislike the center of attention, the main topic of discussion, and a personal crusade.