Long cables to power "ioncraft" to orbit?

The ioncraft is a method proposed for decades for aircraft and spacecraft propulsion:

Ioncraft.

formatting link

It works by ionizing the air by electrical charge thereby creating an air flow between the electrodes, generating thrust. There are several examples of these, called "lifters", made by amateurs:

The Lifters Experiments home page by Jean-Louis Naudin.

formatting link

The problem with them is their power supplies are much heavy than the weight they can lift. But why not leave the power supply on the ground and connect it to the craft by long power cables?

There are carbon fibers that could support their own weight up to hundreds of kilometer of altitude:

Carbon fiber (Dani Eder)

formatting link

And power transmission lines carry electrical power up to 250km away at up to 600 megawatts of power:

Baltic-Cable.

formatting link

This page calculates you can lift 3.91 grams using 7.681 watts of power or about a ratio of 1 to 2:

Lifter Theory.

formatting link

Then you could lift 1,000,000 kg using 2 gigawatts of power. The space shuttle main engines produce a maximum of 37 million horsepower, or

27.6 gigawatts of power:

Boeing: Rocketdyne: Space Shuttle Main Engine Amazing Facts.

formatting link

Then you could leave the large heavy engines and heavy fuel on the ground and use it just to run electrical generators to drive the ioncraft. If the electrical cable was 4 cm wide made of carbon fiber, a 100km long cable would have volume Pi*.02^2*100000 = 125.7m^3. At a density of 1800 kg/m^3 for carbon fiber this would be 226,000 kg. Then twice this number in kilowatts or 452 megawatts would be needed to support the weight of the wire alone. You could have take this from the 10's of gigawatts supplied to the ioncraft or have small versions of the lifter drive all along the length of the power cable itself drawing off some portion of the power to support each small portion of the cable. The question: how much power would be lost by sending it along a 100km long cable?

Bob Clark

Reply to
Robert Clark
Loading thread data ...

[snip]

[snip crap]

Hey stooopid, thrust in this case varies as surface area but payload varies as volume. The demo is bullshit when applied to the real world product.

[snip]

Dumber than a used tampon. Hey stooopid, rockets don't launch straight up - not for long they don't. Getting 100 miles high is a no-brainer. Scaled Composites mostly pulled it off ad hoc. Getting to

25,000 mph at 100 miles altitude is something else again.
--
Uncle Al
http://www.mazepath.com/uncleal/
 Click to see the full signature
Reply to
Uncle Al

Rumsfeld must have taken EST. That was part of the course introduction.

--
Luhan Monat: luhanis(at)yahoo(dot)com
http://members.cox.net/berniekm
 Click to see the full signature
Reply to
Luhan Monat

It's deja vu all over again ;-)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
 Click to see the full signature
Reply to
Jim Thompson

We had a lot of them here. We called them estholes:

"Yes, I'm standing on your face, and I'm wearing cleats, but if you really want to be miserable, it's your problem."

John

Reply to
John Larkin

I assume you will be running two cables for the power, a hot and a return. How about insulation of the wires from each other? Is the conductor here the carbon fiber itself or some metal? Then calculate: a) the total mass of 100,000 ft. of the cables b) the electrical resistance of 100,000 ft. of the the conductors c) power loss through the conductors d) temperature rise through the conductors

"There are known knowns. These are things that we know we know. There are known unknowns. That is to say, there are some things that we know we don't know. But there are also unknown unknowns. These are things we don't know we don't know." -Secretary of Defense Donald Rumsfeld

Reply to
Charles Jean

Choose your voltage. Divide power P by voltage V to get current I. Loss is I^2 * R. You need to work out R using the resistivity of the material and the length. Remember to count both wires.

George

Reply to
George Dishman

Robert Clark wrote: snip childish BS I remember being 13yrs. old and discussing things like this with my roommates at public school. It is a nice exercise being kids, but a grown up person should switch on the probability check before x-posting to all these NGs.

--
ciao Ban
Bordighera, Italy
Reply to
Ban

Some friends of mine who did EST long before I did, already used the term 'estholes' themselves.

--
Luhan Monat: luhanis(at)yahoo(dot)com
http://members.cox.net/berniekm
 Click to see the full signature
Reply to
Luhan Monat

You are assuming that these peopleare grown-up and mature to start.

Reply to
GWBush

If Rummy really said this, then he is much more widely read than I would have believed. This is a variation of a quotation from Lady Burton, attributed as an 'Arabian Proverb': "Men are four: He who knows not and knows not he knows not, he is a fool--shun him; He who knows not and knows he knows not, he is simple--teach him; He who knows and knows not he knows, he is asleep--wake him; He who knows and knows he knows, he is wise--follow him!"

Tom Davidson Richmond, VA

Reply to
tadchem

Umm, if you achieve escape velocity you , err, escape (I.e. you've overachieved)

-- Keith

Reply to
Keith Williams

: >

: >> The question: how much power would be lost by : >> sending it along a 100km long cable? : >

: >Choose your voltage. Divide power P by voltage V to : >get current I. Loss is I^2 * R. You need to work : >out R using the resistivity of the material and the : >length. Remember to count both wires. : >

: >George : >

: I assume you will be running two cables for the power, a hot and a : return. How about insulation of the wires from each other? Is the : conductor here the carbon fiber itself or some metal? Then calculate: : a) the total mass of 100,000 ft. of the cables : b) the electrical resistance of 100,000 ft. of the the conductors : c) power loss through the conductors : d) temperature rise through the conductors

: "There are known knowns. These are things that we know we know. : There are known unknowns. That is to say, there are some : things that we know we don't know. But there are also unknown : unknowns. These are things we don't know we don't know." : -Secretary of Defense Donald Rumsfeld

Makes me think of the little passage from the Koran I think:

He who knows not and knows he knows not is simple, teach him, He who knows and knows not he knows is asleep, awaken him, He who knows not and knows not he knows not is a fool, shun him, He who knows and knows he knows is a leader, follow him.

Eric

Reply to
Eric Chomko

Leaving aside the engineering problems, you have to get to escape velocity to make orbit and at altitude you run out of air to ionise.

-- Stephen Horgan

"intelligent people will tend to overvalue intelligence"

Reply to
Stephen Horgan

Use a uwave beam and onboard rectenna

--
Dirk

The Consensus:-
 Click to see the full signature
Reply to
Dirk Bruere at Neopax

He did.

Yep. Closer to home, note that the first category above is endowed with what we (meaning sci.physics regulars) refer to as "second order ignorance", which is the characteristic of many of our cranks.

Anyway, Rummy did say this and many seemingly intelligent people jumped on this as a "dumb statement", not realizing that it was their own stupidity they were thus proclaiming.

Mati Meron | "When you argue with a fool, snipped-for-privacy@cars.uchicago.edu | chances are he is doing just the same"

Reply to
mmeron

Dirk Bruere at Neopax wrote in news:3f1h43F4q4r6U1 @individual.net:

The working model apparently requires 30 kV to generate the ions.

That implies that you need pretty good insulation on your cables. You need to take that into account.

Neglecting that weight, (assume we can put a few spacers between the wires and their weigh will be negligible, for now) the 452 MW will require 15 kA.

15 thousand amps through the 7.5 ohms (that was from 3 cm diameter aluminum wires, so the weight we need to lift is underestimated by a significant factor, but it doesn't really matter, as you will see) of the wires will drop 113 kV.

The power supply will thus need to put out 133 kV. From this, it should be clear that MOST of the energy will go to heating the wires. The wire will disipate 1.6 GW and 0.45 GW goes for lifting the wire itself. The weight of the payload is negligible compared to the wire.

The wires are going to need to be able to dissipate 254 watts for each 3 cm of length without weakening. I think the 3 cm diameter wire can take it.

Oh, there is one other 'minor' problem. This craft will only work inside the atmosphere because it uses the motion of air molecules to do the lifting.

The lift will fall off with altitude.

So, you don't need to worry about lifting 100 km of wire. It can probably only go to a few hundred thousand feet.

One other minor problem. When it does get up high, it is going to contact some layers of the atmospher that carry high charges wrt ground.

Of course, you may be able to get back much more energy than the lift took.

--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an 
 Click to see the full signature
Reply to
bz

The lifter drive is dependent on the density of the air. It gets weaker at higher altitudes. You could have the drive propel the craft at a higher acceleration than normally used with rockets to get to orbital velocity sooner. Or you could use the lifter drive as a low cost first stage and only carry fuel for the final stage to orbit. Note this is what was also planned for hypersonic orbital craft.

Bob Clark

Reply to
Robert Clark

So it won't work in orbit then.

Reply to
CWatters

How about a 'beanstalk'? Arthur C Clark talked about a structure which is, effectively, a tower which is long enough so that the centrifugal force counterbalances the gravitational attraction. Various SciFi writers have described them. One of the {Red|Green|Blue} Mars series by Kim Stanley Robinson, has a wonderful description of a tether being attacked by terrorists, having it's attach point blown up, and wrapping around the planet at near orbital velocity, destroying everything near the equator.

--
Regards,
  Bob Monsen
Reply to
Bob Monsen

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.