The Dipole Drive: A New Concept for Space Propulsion

https://www.centauri-dreams.org/2018/06/29/the-dipole-drive-a-new-concept-for-space-propulsion/

by sohkamyung • 8 years ago

140 49 8 years ago

centauri-dreams.org

andbberger4• 8 years ago

Huh. I don't see any obvious pitfalls here, the physics and principle of operation are quite simple.

I am struggling to come up with an explanation as to why I haven't heard about this before other than sneaky DOD stuff.

Who would like to bet me a large sum of money that there are currently operation military sats employing this......

EDIT: as epicureanideal pointed out, commenters in the article raised valid points discrediting most of the author's analysis. +/- charged ions get same net kick in the field, comparing proton mass to electron mass is a red herring.

Overall concept still passes sniff test, just needs further analysis. It's going to be all about the angle - you need one ion species to spend more time in the field than the other to get a net thrust

A magnetic scoop with a normal ion thruster might be more practical.

EDIT 2: After further consideration, my concern here is what happens outside the plates. This works for infinite parallel-plate capacitors - you just need to have the field strong enough so one ion species undergoes more net acceleration. In practice the field outside the plates will not be zero and that could spoil everything

EDIT 3: I know nothing. E&M is hard. There will be a current around the outside as you're pumping positive ions to one side and negative to the other. This will induce a magnetic field affecting the ions being pumped through the field. I have no clue what the net effect will be.

epicureanideal• 8 years ago

Some of the commenters on the linked website claim it violates physics principles. Like many people here I took several university physics courses and it seemed plausible to me, but I don't have confidence in what seems plausible to me actually working. There are a lot of "oops, I didn't think about that" types of things when making calculations in physics, from my memory...

Note: After reading post by nickparker here, note that most of the commenters on the linked website who were claiming mistakes with the math seem to be wrong. Nickparker here posted links to what seem to be valid calculations showing the math works out.

andbberger4• 8 years ago

For sure could be subtle crippling problems. But it passes the sniff test with flying colors

ss2003• 8 years ago

It's very good example of why we need real space testing of propulsion concepts. Once we have a bunch of engineers actually living in interplanetary space, where they can do easy experimentation, we will see some rapid developments.

amelius• 8 years ago

I guess the pitfall is that it assumes the presence of a plasma, in which case the question is: can we really call it "space"? But from a practical point of view, that may not matter much, of course.

ballenarosada6• 8 years ago

The presence of a very diffuse plasma in outer space is well established.

sandworm101• 8 years ago

Throwing positives one way and negatives the other? Wont they then seek each other out? Would not this thing create a little storm of ions to negate its thrust?

zamalek• 8 years ago

The cathode ray tube wouldn't work if this didn't work.

namibj• 8 years ago

Still, a propeller works, even though the positive and negative pressure seek each other out.

tbabb• 8 years ago

Ions entering are then propelled from the positive to the negative screen and then out beyond, while electrons are reflected. There are thus two exhausts, but because the protons are much more massive than the electrons, the thrust of the ion current is more than 42 times greater than the opposing electron thrust, providing net thrust.

Does this follow? I thought the force of an E-field on a particle is proportional to charge and field strength-- so the electrons are lighter, but will be accelerated with the same force to a much higher velocity. So wouldn't the reactive force be the same for both?

andbberger4• 8 years ago

You're right, the author's analysis is incorrect. But I don't think this discredits the concept completely.

nickparker• 8 years ago

The particles are passing through a fixed distance (between the screens), wherein they experience a fixed force due to the electric field.

An electron will shoot through that distance in no time flat and accrue very little momentum. A proton will plod through and transfer far more.

For a 250V potential difference with 20m between screens, you can check it yourself here:

https://www.wolframalpha.com/input/?i=(1+electron+charge+%2F...

The ratio of those two values is 42.8.

This also makes it easy to see why heavier ions full of neutrons are _awesome_. They give a ratio of 5343.

ananonymoususer• 8 years ago

Thank you. I was going to post the same concerns about the result being electrons accelerated much more quickly out the front, thus no net propulsion. I am glad I read the comments first.

T-A5• 8 years ago

Put an electron halfway between the screens. It is subject to a force F directed toward the positive screen, so it accelerates in that direction. Once it's through the screen, the force changes sign, but quickly drops to zero (for infinitely large screens, it is exactly zero everywhere outside the space bounded by the two screens). The work done by the electric field is

W = F * d / q

where

d = distance from the electron's start position to the positive screen

q = the electron's charge

All this work is now kinetic energy carried by the electron, so we also have

W = (m_e/2) * v_e^2

where

m_e = mass of the electron

v_e = final speed of the electron

Therefore,

v_e = sqrt( 2 * F * d / (m_e * q) )

If we repeat the experiment with a proton, it accelerates in the opposite direction, toward the negative screen, and its mass m_p is larger, but other than that it all goes as with the electron; the absolute values of q and F (which only depends on electric field strength and q) are the same. So,

v_p = sqrt( 2 * F * d / (m_p * q) )

Divide the two equations to get

v_e / v_p = sqrt( m_p / m_e )

Momentum is conserved, so we also have

M * v + m_e * v_e - m_p * v_p = 0

where

M = mass of the screens (and rest of the spacecraft)

v = speed gained by the screens (and rest of the spacecraft)

and the minus sign accounts for the opposite direction of travel of electron and proton. Use the expression for v_e / v_p to eliminate v_p, solve for v, and find

v = (sqrt(m_p / m_e) - 1) * (m_e / M) * v_e

In numbers, m_p ~ 1837 * m_e, so

v ~ 41.86 * (m_e / M) * v_e

and Zubrin's argument is essentially correct.

grogers2• 8 years ago

Even if there wasn't a difference between electron and proton mass, it would still provide net force if the craft is already moving relative to the background plasma (I.e. swimming upstream). Incoming electrons from the front get reflected at their original velocity, but protons get accelerated out the back at arbitrarily high speed (limited by the electric field). The massive difference in electron and proton means it can provide force even if the plasma is coming from behind, which is much more useful.

dojomouse• 8 years ago

This seems basically the same as an ion engine without the need to BYO ions, relying on the existing ion/electron mix in plasma being sufficient and suitable. Which makes it seem fairly credible. Potential for much higher efficiency than ion engines too if a plentiful supply meant you could sacrifice thrust/ion.

Cathode Ray tube comparison below isn't entirely valid, as the goal of a CRT isn't to generate net thrust.

dnautics• 8 years ago

I would say that a good comparison (of course the math is different, since one is monopole-monopole and another is monopole-dipole) would be saying it's like a continuous gravitational slingshot except with solar wind ions and electricity. With a gravitational slingshot, you use gravity to steal kinetic energy from the larger body (usually the planet doesn't notice); with this you use an electric field to steal kinetic energy from the passing wind.

yellowapple• 8 years ago

If CRTs did generate thrust, though, it would give an entirely new meaning to "our TVs are flying off the shelves!".

ArtifTh• 8 years ago

Looks similar to Ionocraft[0], but only for low density plasma instead of an air.

[0] https://en.wikipedia.org/wiki/Ionocraft

tigerlily• 8 years ago

I guess there's only one thing left to do. Experiment!

ridgeguy• 8 years ago

Yes.

Seems like this would be a fairly straightforward proof of concept experiment to fly outboard on the ISS. Basically a couple of charged screens with force transducers on the screen mounts. It would seem this is within the bounds of Figure 1, The Dipole Drive Accelerating Within a Magnetosphere.

Wish this paper had been published a year or two before SpaceX launched Elon's Tesla. They could have tested a new interplanetary drive.

acertainfool• 8 years ago

Where are these charged screens coming from? Is it possible to just magically have charged screens that remain at the same charge regardless of all interactions or am I seeing a place where someone has to bring their own ions?

ballenarosada6• 8 years ago

This won't work. The basic idea about the protons gaining more momentum than the electrons is valid. But the dipole creates an opposite field outside of the charged plates. Protons will be decelerated until they pass the first, positively charged plate, then accelerated through the plates, then decelerated back towards the negative plate.

This is all clear if you consider the ions falling through a potential field. The potential is 0 at infinity, positive at the first plate and negative at the second. An incoming ion starts at 0 potential, climbs a big hill to get through the first plate, then falls down below 0. Then on the way out it has to climb back to 0 potential at infinity. So the ions gain energy inside the plates but lose it all back on either side.

andbberger4• 8 years ago

Thanks for this, excellent simple argument cutting right to the core of it.

I'm a fool. Don't speculate about E&M before you've had your coffee kids.

T-A5• 8 years ago

https://physics.stackexchange.com/questions/12309/electric-f...

ballenarosada6• 8 years ago

The electric field outside an infinite capacitor is zero. For a finite capacitor, there is a nonzero field. The importance of the infiniteness assumption can't be understated--such a capacitor cuts the universe in half, and every point of one half has the same electric potential.

On the other hand, if the capacitor is finite, then the surface integrals over the plates are not equal.

gus_massa8• 8 years ago

I almost believe in this ... but no.

It's just a parallel plate capacitor, with holes in the plates that allow the surrounding plasma to pass trough. The question is:

Why a plate capacitor with holes in the plates inside a diluted plasma is not a perpetual mobile?

Let's assume that initially the device is with speed 0 inside a huge bath of plasma that is also at speed 0. We charge the two plates of the capacitor instantly and magically. In this model and the calculations the ions never hit the plates because they have holes, so the plates never discharge. There is some energy in the electric field of the capacitor.

Now the device is operational and it starts to move, the ions also move. Now the devices and ions have some kinetic energy.

* If we believe the whole idea the ions just go away, and the plates never discharge, and new plasma enters the device, so it continues to accelerate forever. So the total energy of the system increase and increase, that is something impossible.

* If we don't believe the whole idea, the problem is that the ions accumulate outside the plates in some kind of ion cloud, even if they never hit the plates. Initially the effect is small, but eventually the charge of the ion clouds compensate the charge of the plates and the plasma inside the device feels no net force, so the device moves at a constant speed. Essentially, some part of the energy that was stored in the electric field was transformed into kinetic energy, but there is only a finite amount of it, so the device must reach a maximal velocity.

---

There is some complications, because in the Solar system the plasma is not static, because there is solar wind. So, just imagine that you pick a reference frame (i.e. starship) that is moving at the same speed of the solar wind. In this reference frame (i.e. starship) initially the plasma looks static and the giant capacitor is moving. When it is turn on, there is some energy in the electric field and some kinetic energy. After a while, the charges are redistributed, and the speed of the giant capacitor changes, but only a finite amount, it can't continue accelerating forever. If you see the scene from the normal reference frame (i.e. the Sun) the device is initially static and after a while it is moving with a cloud of ions around it, but the total change in speed is the same.

I guess there is some drag, and the final speed in the starship reference frame is 0 and the final speed in the Sun reference frame is the same of the speed of the solar wind.

---

About other comments:

* I agree that the momentum transferred to the protons and electrons is different. With some assumptions the ratio is 43, but in a moving device in a moving solar wind the calculation is more complicated. The exit speed of the electrons is 100 times faster than the solar wind, so it makes almost no difference. The exit speed of the proton is roughly the speed of the solar wind, so the calculation must be fixed, but there is a net effect that is no 0. (Until the ions cloud accumulate enough charge.)

* In an ion thruster the first step is to ionize the gas. Here they start with an ionized gas. The difference is that in a ion thruster they separate the positive and negative field.ions, so they waste some energy in the separation and then they accelerate only the positive ions. The separation of the ions is an active process, not just a static electric

* Hat tip for sandworm101 for the comment https://news.ycombinator.com/user?id=sandworm101

---

About the article:

The acceleration of the electrons is a form of drag, which is provided for by loss of spacecraft kinetic energy. It therefore could, in principle be used to generate electric power, partially compensating for the power consumed to accelerate the protons.

This makes no sense. The acceleration of the electrons create some thrust in the wrong direction, but it's not like a drag, it's like a thruster in the wrong direction.

To see what the performance of a dipole drive might be, let us work an example, assuming a 500 W power source to drive the system. The electron current negates about 2% of the thrust (1/43rd) produced by the proton current. The maximum possible jet power is thus about 490 Wj. Assuming additional inefficiencies, we will round this down to 400 Wj, for a total system electrical to jet power efficiency of 0.8. The relationship of jet power (P) to mass flow (m) and exhaust velocity (c) is given by: P = mc^2/2

I'm not sure that mixing thrust and energy is a good idea, but anyway they assume that the efficiency is 80% and use the usual formula to get the speed. The problem is that assuming that only a 20% of the energy will be wasted is optimism, not a real calculation. How much energy is lost due to the collision of the ions and the plates? How much to recharge the plates? How much is lost in real drag? How much is lost as heat? ... There are a lots and lots of factors, and then you have to consider the nasty ions clouds.

grogers2• 8 years ago

It's not a perpetual motion machine because it requires energy. Every proton that gets accelerated lowers the charge on the capacitor plates by that same charge. So you have to supply current to keep the plates at a constant voltage.

gus_massa8• 8 years ago

If you look at the Figure 1 "The Dipole Drive Accelerating within a Magnetosphere." it clearly says "Accelerated proton" and "Reflected electron" and it shows how they pass thru the device. They are just accelerating the plasma particles, without any absorption or emission from the plates.

adrianmonk3• 8 years ago

Why are you even talking about it as a perpetual motion machine? I don't see any claim anywhere in the original article that it is. He even talks about nuclear reactors used to power it.

You're refuting a claim that doesn't even exist.

And in doing so, you're also assuming the particles around it have "speed 0" when the explanation clearly describes individual particles as having non-zero velocity so that they can enter the space between the two plates on their own, as it is claimed that there would be no net electric field outside of them.

gus_massa8• 8 years ago

They never claim it is a perpetual motion machine. They just invented a machine and if it works as intended it would be a perpetual motion machine, so we know it will not work as intended. [1] They don't claim that it is a perpetual motion machine, if you analyze the consequences of their claims you conclude that it is a perpetual motion machine. I guess they didn't notice this, it's not an obvious consequence in the usual reference frame.

In my description I change the reference frame because it's much easier to do the calculations. In the reference frame that moves at the same speed of the solar wind the plasma has speed 0, but it doesn't mean that all the particles have speed 0. It means that the number of particles that are moving in one direction is (almost) equal to the number of particles in the opposite direction. [2] It's like the air inside a quiet totally sealed room that is left alone for a few days. The air "doesn't move" but the oxygen and nitrogen molecules are moving very fast. The speed that they have is related to the temperature. But in bulk they don't move and if you put some small pieces of paper or strings they will not move (unless they are really tiny). If you have some container like a glass inside the room the molecules that are inside the container will change constantly, but the number of molecules inside the container will (almost) no change. The problem in my version of the experiment is not that the device will get empty.

[1] Assuming we don't need to change the current accepted laws of physics to explain this. It would be even more groundbreaking.

[2] There are more accurate technical description but this is good enough.

DennisP• 8 years ago

It's not a perpetual motion machine, for the same reason a jet engine isn't one: it has an energy source, and it gets reaction mass from the surrounding environment.

gus_massa8• 8 years ago

They have a energy source, but they actually not using it. They need the energy source to charge the giant capacitor.

But just imagine that after it has been charged to the nominal value the energy source get's broken, and as an emergency measure you just blow up all the conductors and make the two capacitor plates fully isolated. Does the device continue to work? For how long? How do the capacitor discharge?

In their description the particles never hit the plates, so the plates don't discharge. In the real word it's more complicated, but it's an usual assumption for this kind of problems. The real word has these nasty details that reduce efficiency, but they may obscure the main ideas and problems.

If you still are not convinced, just make a hole in both plates and put a isolator tube between the holes. Also, put some isolator around the plates. Make all the plasma only go throw the tube.

Now the calculations are more difficult because it's no longer the electric field of a perfect plate capacitor, but any finite capacitor has problems at the border. If the hole is small enough the using the formula of a plate capacitor is good enough. Let's see a graphic, were the double lines are the insulator and the +++ and --- are the plates:

               entry
  ╔══════════════╦ ╦═══════════════╗
  ║++++++++++++++║ ║+++++++++++++++║
  ║              ║ ║               ║
  ║--------------║ ║---------------║
  ╚══════════════╩ ╩═══════════════╝
                 exit

The efficiency will be abysmally low, because you will have a huge starship with a tiny motor, but the important detail is that the effect will be there. Do the plate discharge?

Moreover, in their calculation they assume that the 80% of the energy is transfer to the kinetic energy of the protons. Where does their calculation change in the model? The efficiency must be also 80%. (Perhaps as efficient, but with a smaller thrust.)

abakus• 8 years ago

Talk is cheap, show me the prototype?

The Dipole Drive: A New Concept for Space Propulsion

Related Stories