How zero-exhaust rocket to stars works.

How Billy T’s Gravitational/Electric (No propellant) rocket, Bootstrap I, works (As in pull your self up by your bootstraps):

Bootstrap I leaves Earth with chemical thrust but once in deep space, the nose section begins to spin and centrifugal force unfolds a large aluminum coated (front side) cloth circle perpendicular to the axis and flight path of Bootstrap I.

For purpose of simple numbers in the explanation, I assume that protons (sometimes called hydrogen nuclei) are 2000 times more massive than electrons and I ignore relativistic effects. Bootstrap I throws 2000 electrons out the front of the rocket on its axis with half the speed of light for every proton thrown out the rear, also with half the speed of light. Thus, the expelling of charged mass does not produce any net thrust, but does charge Bootstrap I strongly positive wrt the electrons out in front and slightly negative wrt the trailing Proton Cloud, hereafter referred to as the PC (and the leading Electric Cloud as, the EC). There is of course an electric field between PC & EC with polarity such that the positively charged Bootstrap I is attracted forward towards the EC and slightly pushed forward by the PC.

Because the mass of the PC and EC are exactly equal but the individual protons have 2000 times more kinetic energy than the electrons they travel farther from the rocket before their motion is arrested and they begin their trip back to the rocket ship. Thus the rocket “falls” towards the gravitational field of the nearer EC more than it is retarded by the gravitational field of the more distant PC. So Bootstrap I is really a dual-thrust hybrid, using both gravitational and electric forces for propulsion.

The electrons continuously ejected to form the EC are more rapidly slowed by the stronger front electric field. All fall back towards the rocket and most are colleted on the nose disk, but their mutual repulsion in the denser EC causes some lateral spreading a few miss the nose disk. Most of them eventually find one of the protons trailing in the even more positive PC behind. Once there, they form hydrogen atoms and are lost to space.

Fortunately, the loss rate can be controlled. (It decreases with the charge ejection rate and also with the speed of ejection as with lower ejection speeds, the charges turn around closer to the rocket. (Effectively the disk is ‘bigger” with greater recapture efficiency.) The loss rate is automatically adjusted to equal the “capture rate” (Space does have low density neutral mix of protons and electrons, which are sweep into the nose disk as the rocket travels thru space.) Ironically, the denser the cosmic cloud the rocket is passing through, the faster the rocket can accelerate, sort of a “negative friction” effect.

All of the protons that were not neutralized to hydrogen by the electrons, which missed the nose disk, also return to the rocket ship, primarily by first striking the back side of the of the slowly spinning disk, where they are neutralized to hydrogen by the returning electrons and soon form hydrogen molecules. Because of the near absolute zero temperature of the disk, this hydrogen is a liquid film, which naturally aggregates under the influence of surface tension to form liquid drops, much like dew drops form on cold surface from water vapor molecules in the air. The “undulation pump” (ripple distortions propagating towards the axis in the disk) returns this hydrogen to the storage tank.

Bootstrap I was 1% of the way to nearest star (Barnard’s Star) when disaster struck. I can not go into details just yet as the patent claims for the improved Bootstrap II have not yet been filed.

Perhaps you might like to speculate what went wrong.

 

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Lol

 

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Hmmm... I've going to venture a (pretty confident) guess that such a device won't behave in the manner you describe; specifically that if all the electrons and protons return to the ship there will be no net change in the ship's motion.

I expect someone with more skills could demonstrate this elegantly using properties of conservative fields or something, but I'll stick with crunching out a 1-D equation of motion for the ship, based on throwing out two single equal masses with different and opposite charges in opposite directions.

I'm going to ignore gravity. Its practical effect would be negligible anyway.

 

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I think you understood, but just to be more clear, there is alway an excess of electrons separated from the ship leaving it always positively charged. If your dynamic analysis shows that the protons take much longer to return to ship (They went farther away and the electric field returning them is much weaker, so that surely is the case.) then throw out more than 2000 electrons per proton ejected so that the EC has more charge than the PC, the rocket is always positively charged etc. Also note that the + and - charges need not leave at the same speed, all that is required is that their ejection momentum is zero or to the rear.

I.e. set up your analysis in parmetric form, not with my numbers.

 

Billy T,

Question. Why would the charge of the EC field be stronger than the PC field? An individual proton has 2000 times the charge in electron volts (actually 1836) as in individual electron. Wouldn't the total charge of both fields be the same?

Billy T,

Another question. Why would the individual particles, which are initially travelling at .5c, not escape the magnetic field of the ship quickly? The magnetic field would be very weak and short-ranged wrt the ship. Since all the particles travelling each stream, i.e. positive stream and negative stream, are of like charges, they would not attract each other, but repell each instead.

Billy T,

I doubt much of a 'gravitational field' can be generated by the tiny mass of the protons or electrons.

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An initial speed of .5c should place the particles well above the 'escape velocity' of the ships gravitational field, don't you think? Are you 'crank-fishing' with your gedanken, Billy T?

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Hi 2inq,
A proton has an equal and opposite charge to an electron.
An electron-volt is a measure of mass or energy.

 

Pete correctly answered your first question for me. Your second and third questions assume there is a magnetic field, which there is not. (The charged particles return to the ship because of electric forces.)

The last question answer is "No." I posted in "Propellantless propulsion, apparently" thread, even before James R did to state it was a silly claim/ system. If one wanted to make such a claim, I thought one could do better than that microwave nonsense - so I did and invite all to tell why my Bootstrap I will not work either. I like to teach and think I did a much more plausible job of describing how to make PP, "Propellantless Propulsion," not urine but there is some degree of similarity - don't you agree?

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Billy T,

Yes, I knew that already, but for some reason I drew a blank when asking the 'total charge' question. I think I might have been confused from my reading of research on lightning strikes by NASA, etc. The positively charged lightning strikes emitted from charged clouds are magnitudes more powerful than negatively charged strikes from clouds. I'm still not sure why.

Billy T,

Billy T, wouldn't a charged particle moving at a constant velocity (.5c) produce a magnetic field? Don't electric fields lose attractive force at the inverse square ratio, same as gravity? I'm not certain at the diminishing force ratio of a magnetic field, anywhere from inverse square to inverse cube, correct? Each individual proton or electron would soon have an almost non-existant force between itself and the rocket. Notice I did state almost, not totally non-existant. The attractive current (force) would be transferred by an electromagnetic particle, a photon, correct?

2inquisitive,

Billy T,

I'm not sure if your 'no' answer is in response to this question, or the question about 'crank-fishing'. I assume you skipped this question and were stating your example was the use of a strawman to attack the 'EMdrive' concept.

 

Yes, Yes, & Yes, but more like inverse cube at reasonable distances. However your are forgetting that the charged particles slow as they go out, stop and return, so there is no net current to make a magnetic field. True the electric field is not strong when charges are well separated, but conpared to zero, it is big. It is true that 0.5c may be so large that the ions at least do escape so the ejection velocities may need to be much lower. It gets complex as if many are "out there"
the electric field could be come large. - I just made up numbers, as I am lazy, but Pete said he intended to do it correctly.

My "No" was answering your 'crank-fishing' question. I attacked the "EM drive" in the original PP thread. I thought it so silly that I could do better in sense of make more convencing nonsense, but you and others will be the judge of that. I did skip your "gravity is weak - is it not?" question as Pete had already told he would ignore gravity effect compared to the electric one.

 

Wake up and smell the field lines, Billy. An individual charged particle will provide a field intensity diminishing as the INVERSE SQUARE of its distance to the test charge, according to all the accepted electrodynamic theories that I have read.

Sorry, I forgot for a moment that you share the allergy to actually reading science texts that other prominent posters suffer from.

 

Thanks, Billy T, for your answers.

With respect to the 'Propellantless propulsion, apparently' thread, I think there are some misconceptions as to what is actually proposed. Even the EM-drive moniker is misleading, although I suppose it was named as such by Shawyer. I agree there will be no acceleration or 'propulsion' generated by the device itself. I also think that the tests that preported to measure a reduction in gravitational mass are a distint possibility. I think the best way to understand the resonat cavity results is to use an analogy concerning inertial mass itself. The microwaves inside the resonating cavity supposedly produce more force on the large end of the cavity than the smaller end. I think an analogy can be used that the force produces a preferred direction for inertial mass, changes inertial mass to a vector quantity. By the use of an outside force, the cavity is easier to move in one direction than all others. It is as if there is a weight reduction in that vector. The 'EMdrive' will not move on its own in that direction, but any forces from thrusters to gravitational attraction will 'see' the reduction in inertial mass. In Earth's atmosphere, for instance, if the difference in the magnitude of the forces at each end of the cavity were equal to the surface weight of the cavity, the Earth's atmospheric pressure would push the device upwards like a helium balloon is pushed up. The atmospheric pressure does the pushing, not the force generated inside the cavity. While in orbital freefall, I don't think the device by itself can be used to raise orbital position, but it could make it possible that a small thruster, such as an ion drive, could easly push the satellite the cavity was attached to higher.

 

A short coupled magnetic dipole may provide a variety of field strength intensity variations in addition to the inverse square solution. The inverse cube solution is correct for a very specific and indeed perfect circumstance for a point particle commanding a magnetic field.

 

Seems to me that if the source of protons and electons is hydrogen, and 2000 electrons are being expelled for every proton, then there will be 1999 protons left on board for every proton ejected....

This means that the spacecraft would be strongly positive compared to the trailing proton cloud.

Thus, the pc would be repulsed by the spacecraft. Not attracted.

Don't know what this would do to the craft's motion, but it would mean that hydrogen fuel would be used up at a greater pace than is being posited at present.


Edit:
Also. As the craft progressed, the positive charge retained in the body of the vessel would grow larger and larger. Thus, as time passes, the energy required to eject electrons would increase. Although, this might be complemented by less energy to eject protons.

Of course, this building positive charge could be reduced from time to time by a conversion to a proton drive where the electrical charge of the vessel would be equalized. This, of course, still leaves us with the problem of hydrogen fuel being used up at a greater amount than posited.


Edit again:

Hmm. Perhaps the natural behavior of the vessel would be an oscillation?
That is, the ship's positive charge would push away the PC at first, and cause the EC to pull closer and closer to the ship. Until, finally, the EC is basically at the ship and the electrons can't be ejected at all because of the immense positive charge of the ship.

At this point, the electrons would 'outweigh' the protons because of the protons lost to the PC. Now, the ship would have a negative charge and would perhaps start tugging on the PC (assuming it's not too far away for such an effect) and would thus go backwards.

A seesaw.
Back and forth until the hydrogen runs out due to entropy (or whatever.)

Could this be the disaster you spoke of?

 

We believe that the perveyor of this astonishing theory is bursting at the seams to reveal the mathematical calculations which clearly prove that a net unidirectional thrust is provided.

Show me the math!

 

I never stated that no magnetic field is created by a linear current. What I stated was that there was no net current as all the charged particles leaving also return. This is very similar to why there is no magnetic field outside of a coaxial cable carrying DC one way on the central conductor and the other way on the surrounding "sheath."

I sold almost all my physics books when I moved to Brazil as my baggage was limited and I had lost interest in physics compared to my interest in how the mind functions and what it is etc. I also do not Google much, but rely on memory (I have read many physics books and "yards"* of physics journals, even written a few papers for them, but not many as most of my work was classified. I hold several "secrete patents" - bet you did not know they even exist, but it is no secrete that they do.)**

If you want to consider the magnetic field near a long straight current wire ("near" = far from the return current path) I am reasonably sure it falls off more slowly than inverse square. I seem to remember inverse log but also and conflictingly that the line integral around wire on circular path must be constant so that, if true, would force inverse linear drop off. - It is a very simple standard problem that I must have included on exams many times. - I just forget the answer and am too lazy (and too tired just now) to work it out, but nearly sure (from memory) your inverse square is wrong.
--------------------------------
*perhaps that is "meters" of journals

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** You can not offer to sell them. One of mine, I can not even tell its name/ title. All were developed under US Navy support. As I understand it, they exist to protect the government from any posibility that they might need to pay royality to some company that also makes the invention later with company funds.

 

What part of the phrase "...there is a magnetic field, which there is not." is it that I don't understand?

 

I am not going to be engaged in a peripheral arguement about the field intensity fall-off of magnetic fields. There are too many possible scenarios and almost any particular one that is brought under discussion can be proved correct.

In terms of the field intensity between charged particles in the absence of an effective magnetic field, it is the inverse square.

You can take it from me

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.

 

Good. I never mentioned magnetic fields (as there are none) in the early posts. - YOU first mentioned them, wrongly stated their fall off, and continued to think them important!

I only tried to correct your error about magnetic dipole fields existing and how they decrease with distance - not be because magnetic fields have anything to do with the subject of his thread, only because I always try to correct errors I notice, no mater how far "off thread" they are.

Take the following pledge with me:

I will not discuss magnetic fields more in this thread.

 

I imagine the disaster is that you could not in reality continue to throw out 2000 electrons for every proton, as if you did the charge on the spaceship would grow large without bound. In reality what happened is that the flight time of protons lengthen and electrons shortended increasing the total charge pc and decreasing charge ec. The charge sc (spaceship charge) would also be affected and the problem is very complex, but when dynamic balence is reached the net force on the system sums to zero and the rate of proton emission and electron emission will be equal.