Author Topic: Zubrin's Dipole Drive  (Read 3775 times)

Offline Proponent

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Re: Zubrin's Dipole Drive
« Reply #20 on: 10/13/2018 09:06 PM »
I get:
0.1* 2.66e-26+0.9*1.67e-27 = 4.163e-27 kg/ion using his 10 % oxygen, 90% proton values. Multiply by a million ions per cm^3 to get mass density. Estimating flux I get

4.16e-21 kg/cm^3 *7800 m/s *pi*(80 m)^2 in mg/s = 0.652 mg/s Which is exactly 10 times bigger than the number he claimed. It seems like there is a typo somewhere (maybe on my end)

That's what I get.  So if his drive works at all, it ought to work quite a bit better in Earth orbit than Zubrin says.

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His quadrupling of the voltage, while cutting the radius in half, and getting half as much thrust leaves me confused, especially as he somehow claims constant power.

Well, with half the radius, presumably the ion flux will be reduced to a quarter.  If the energy per ion is constant, then quadrupling the voltage would then maintain constant power.  Quadrupling the voltage should quadruple the kinetic energy each ion picks up passing through the thruster.  If the kinetic energy of an ion were negligible before it reached the thruster, then quadrupling its energy gain would double the exhaust velocity.  So, with one quarter the mass flux and double the exhaust velocity, the thrust would be halved.  The initial energy of the ions isn't completely negligible (for protons, I get that 7800 m/s corresponds to 5.0e-20 J or 0.31 eV), but it's not too large.  Still, I don't see why the power consumption relates to ion energy.

Offline ChrisWilson68

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Re: Zubrin's Dipole Drive
« Reply #21 on: 10/13/2018 10:06 PM »
For Zubrin's drive, where is the energy being put in?  What charges are being moved?

A power source on the ship charges the grids. Ions in the plasma wind are accelerated by the charged grids. The reaction mass is the plasma wind. Zubrin describes it in the video.
This doesn't change the argument that the particles coming and going have the same potential energy as they do at infinite distance, so the net force must go to 0. There should be a way to make a related concept work, but it would take something more complicated. After initially charging a capacitor, it takes no energy to maintain it (under ideal conditions, with no leakage) so the lack of continuous energy input needed is another reason it is clear you need something more complicated if you want to make this work.

Exactly.

Reaction mass isn't the issue.  The idea is clearly that the particles of the interstellar medium end up being accelerated.  So there's no violation of conservation of momentum.

But you need something more than fixed (relative to the ship) electrostatics or you violate conservation of energy if there's a net acceleration of the interstellar medium particles.

Good point. Particle accelerators alternate electric fields at the appropriate time to accelerate charged particles, but the dipole drive doesn't do that.

What about the claim the fields cancel each other outside the two charged grids? There are obvious problems with that since the charged grids are not infinite, but would there be enough of an effect directly in front and behind the direction of travel to create some net thrust?

Approximating it as infinite planes of charge is only valid for particles close enough to the planes that edge effects aren't significant.  No matter how big the grids are, though, there's going to be some distance that's much larger than the size of the grids, and at those distances the edge effects dominate.  Since the interstellar medium particles start out at such distances and eventually end up at such distances, you can't ignore the edge effects.

It's the edge effects that mean that the potential energy far in front of the craft and far behind it are the same, even if that wouldn't be true of truly infinite planes of charge.

As the particle is going from very far in front of the craft to the region close to the first plane, it will be decelerated by these edge effects.  This will happen slowly over a large distance.  Then, it will accelerate quickly between the grids.  Then, as it goes from close to the tail grid to an extreme distance, it will again decelerate slowly.  Conservation of energy says that the net of all these accelerations and decelerations must be zero.

Offline edzieba

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Re: Zubrin's Dipole Drive
« Reply #22 on: 10/15/2018 01:24 PM »
As the grids are acting as a charge separator (electrons fired forwards, protons and cations fired rearwards), there should be some non-equal effect from the positive particles being fired at the positively charged 'exhaust'.

Offline Proponent

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Re: Zubrin's Dipole Drive
« Reply #23 on: 10/15/2018 06:22 PM »
OK, after reading Paul Gilster's article on Centauri Dreams, I think I understand where the mass-flow rate of 65.2 μg/s comes from  Taking, like Zubrin, the example of a thruster with a 500-W power source, Gilster assumes an 80% efficiency of conversion of electrical power into jet power.  Assuming, further, a grid potential of 64 V, Gilster calculates a jet current of (400 W)/(64 V) = 6.25 A.  Since a proton's mass-to-charge ratio is (1.67e-27 kg)/(1.60e-19 C) = 1.04e-8 kg/C, that current corresponds to a mass flux of (6.25 A)(1.04e-8 kg/C) = 6.51e-8 kg/s, which is what Zubrin "requires."

The grids' charges will tend to leak away over time.  Counteracting that will require electrical power, but why that power would be approximately equal to the jet power is a mystery to me.

Offline Proponent

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Re: Zubrin's Dipole Drive
« Reply #24 on: 10/15/2018 06:38 PM »
As the grids are acting as a charge separator (electrons fired forwards, protons and cations fired rearwards), there should be some non-equal effect from the positive particles being fired at the positively charged 'exhaust'.

I would think that wouldn't be too much of a problem as long as the speed is much less than the exhaust velocity.  Then the density of ions behind the thruster should be low.

Online KelvinZero

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Re: Zubrin's Dipole Drive
« Reply #25 on: 10/17/2018 01:41 AM »
(EDIT- when we talk about canceling out, we should be careful whether we're taking about fields or potentials. The fields cancel, but it leaves a non-zero potential behind.)
This is me refreshing my school level physics rather than advanced topics :)

Let me see if I have this right:

We can disregard forces and just look at potentials. If a particle moves from one potential to another, it gets work done on it. The path does not matter.

With finite-sized planes, distant particles both in front and behind are at almost zero difference in potential.. so you would somehow have to do work to push them into the near-center volume where forces approximate two infinite planes. Whatever these fringe forces are, they must somehow cancel out what you gain as they cross the gap.

Two ways to make this work:
(1) Turn the plane's charge on and off.. I think I can see how this now consumes energy because it takes work to separate those charges and you don't get quite as much back after particles in space have had work done on them by this configuration.
(2) neutralise the expelled protons so that all those fringe forces become irrelevant at least on the departing side. Now you are back to doing straightforward work again, collecting electrons and moving them against the force of the planes, in order to neutralise the protons on the other side.
« Last Edit: 10/17/2018 08:17 PM by KelvinZero »

Offline Proponent

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Re: Zubrin's Dipole Drive
« Reply #26 on: 10/17/2018 05:26 PM »
Two ways to make this work:
(1) Turn the plane's charge on and off.. I think I can see how this now consumes energy because it takes work to separate those charges and you don't get quite as much back after particles in space have had work done on them by this configuration.

It will take energy to charge up the grids, but you get that energy back when you discharge them, minus some inefficiencies.  It's just a capacitor.  You can charge up a capacitor and then extract the energy later.  I don't think pulsing solves the problem.

I think having different ion densities inside and outside the thruster would do it, if you could get the density difference in a way that didn't involve moving ions through the electric potential.  A conventional ion engine is an extreme example, as is an ionocraft.  In both cases, the density within is higher than the density without, though that's not the only way in which they differ from a dipole thruster.

It would be easy to arrange a lower ion density with in the thruster: just seal it off and don't allow any ions in.  Then it will work as a brake.  But it would be a better brake without the negative grid, i.e., if it were a Jahunen electric sail.

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(2) neutralise the expelled protons so that all those fringe forces become irrelevant at least on the departing side. Now you are back to doing straightforward work again, collecting electrons and moving them against the force of the planes, in order to neutralise the protons on the other side.

OK, this I can sort of see.  Incoming ions must still climb the upstream potential hill, but departing neutralised ions (atoms) need not, so you can see why there would be a net thrust, albeit of half the magnitude claimed by Zubrin.  And, as you say, we do work in collecting electrons from the positive grid and moving them to higher potential on the negative grid and then shooting them off into the exhaust beam, so it no longer seems like we're getting something for nothing.
« Last Edit: 10/17/2018 06:34 PM by Proponent »

Offline Proponent

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Re: Zubrin's Dipole Drive
« Reply #27 on: 10/17/2018 05:58 PM »
Put an ion gun (i.e., small conventional ion engine minus the neutralising electron gun) aboard, and shoot ions off (might as well be rearward, though thrust isn't the main point) to put a negative charge on the craft as a whole while maintaining enough of a net positive charge on the forward grid to attract the electrons that will be needed to neutralise the exhaust.  This reduces the potential hill that incoming ions must climb.  Given that electrons are much lighter than ions, they tend, at the same temperature, to zip about quite a bit faster than ions, so we should be able to operate at a somewhat negative overall potential while still attracting as many electrons as ions.  It's well known, for example, that the solar wind tends to put net negative charges on spacecraft, because of this effect.

I'm thinking out loud here and anticipate that I've made some stupid mistake.  Please put me out of my misery by telling me sooner rather than later what that mistake(s) is!

Offline Proponent

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Re: Zubrin's Dipole Drive
« Reply #28 on: 10/17/2018 08:08 PM »
Actually, it seems dumb to have both an electron gun and an ion gun. Maybe omit the ion gun but don't quite fully neutralise the exhaust, leaving the vehicle with a net negative charge.

Online KelvinZero

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Re: Zubrin's Dipole Drive
« Reply #29 on: 10/17/2018 10:25 PM »
Actually, it seems dumb to have both an electron gun and an ion gun. Maybe omit the ion gun but don't quite fully neutralise the exhaust, leaving the vehicle with a net negative charge.
I was just thinking the same thing. Suppose you had a big negative charge that pulled in protons without actually stopping them (eg like that polywell fusion design) , then if you pointed an electron gun in one direction, roughly matching the speed of the outgoing protons then it would be a bit like a rocket nozzle where you are containing an explosion in a box but you don't feel force on one side because you took the wall away.

That particular design probably creates a lot of drag though. Also you have to replenish your electrons somehow. At least collecting electrons is probably a lot easier than collecting protons.

Online KelvinZero

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Re: Zubrin's Dipole Drive
« Reply #30 on: 10/18/2018 11:57 PM »
@momerathe: yeah I think you're right, that's pretty much my thinking as well. This is also why I'm more intrigued with a related idea I posted in a thread a while back: https://forum.nasaspaceflight.com/index.php?topic=46146.0
Hi I just got around to looking at the associated PDF. It actually mentions the Zubrin idea and brings up pretty much the same issues that people have on this thread (a static finite system won't produce thrust, potential goes to zero except near the device etc)

Apart from that I have not really understood it yet, except that it is similar but something oscillates.

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