Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1875427 times)

Offline PaulF

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One way I came to grips with accelerating inertial frames was I realized what happens in between two frames of special relativity.  For there to be contraction of outside space, then, when when one is accelerating is when the contraction happens.  I was thinking of how relativity seems to enhance classical effects like momentum, mass, and energy and I wondered if us living in an accelerated frame (gravity) means our frame is continuously contracting and relativity enhances this?  That is, space continuously contracting into the earth suggests a flow into the earth.  It sort of made sense that something was dragging us as it flows in. 
I'd like to kick in on this a little. After reading this my brain went into auto-though experiment mode.

Let's say space is continuously contracting in the presence of gravity. Relatively speaking, that means it could also be the universe expanding around it. Now I am no expert on this subject, but I know that is what we are observing today, expansion of the universe. What if the expansion of the universe is not due to expansion, but due to contraction of the observers (us + observational equipment, i.e. hubble etc)? I can't think of a method to prove which of the two it is, or is there a way?

P.S. I just saw Warptech mentioned this very same thing. Maybe a discussion is in order :)
« Last Edit: 06/07/2015 11:25 AM by PaulF »

Offline Rodal

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Absolutists? Count me among them. Conservation of energy and the isotropic universe are some of the founding principals of modern science. Sure, anisotropic explanations might be true, but as EMdrive proponents have largely conceded, the operation and design of the EMdrive crucially depends on the violation of one of these principals. Would you refer to >99% of physicists and other scientists dogmatic absolutists? They, and for that matter journal editors, R1 PIs, and governmental program managers in places like the NSF and DOE, aren't going to be convinced that there is a problem with those assumptions unless there is some really serious experimental evidence.

As you are new in the thread (welcome  :) ) and from your assertions it appears you are unfamiliar with the subject matter being discussed in my post that you quoted.  Suggestion: search for the peer-reviewed papers of Bart van Tiggelen et.al. and read about chilarity.  You may find them posted by @Mulletron, in the last hundreds of pages of this thread.
« Last Edit: 06/07/2015 12:36 PM by Rodal »

Offline SeeShells

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Really a pleasure to read posts on all sides of the em drive...wish I were in the 0.5% club like many here ;)

Since I tend to go macro...I am visualizing fields whose strength is inversely proportional to spacetime...iow...weak forces can span huge distances. Pehaps we are leaving our preoccupation with strong nuclear forces and heading towards the understanding and utilization of the opposite.  As always, thanks for activating some of my dormant brain cells  8)
Sometimes I think of the QV of space as one giant Quantum Entanglement. Big enough macro?
shell

Offline WarpTech

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Why are you veering off into abstractions about GR? You seem to be obsessed with it. You haven't addressed head-on my simple observation that you have presupposed that energy conservation obtains. In an early post to this thread
http://forum.nasaspaceflight.com/index.php?topic=36313.msg1369875#msg1369875
I showed that if you assume P = F v,  then conservation automatically follows.

I am using Newton in a field-free flat spacetime at relative velocities severely smaller than c. There is no GR here. There is no SR here. There is classical mechanics and you have just made a postulate that guarantees that conservation obtains. The price you pay is to have selected a preferred frame.

The ghost of Einstein is going to hunt you down.

Look, you were arguing with me about a preferred frame 10+ years ago. Yes, in Newtonian gravity v<<c it implies a preferred frame. You must use GR and "understand it" in order to eliminate that implication. You're not doing your homework to understand GR well enough, or as I've explained it, to get past this and it is not my job to convince you. I've found trying to convince absolutists of anything is not worth the effort. An object in an inertial frame is not equivalent to the same object with a higher inertia content in another inertial frame in GR. It is not a Lorentz Transformation anymore once matter is accelerated and it's inertial mass has changed. The inertial frames are "different", not "identical" because the inertial mass has changed and in so doing, space-time was affected.

Now, how do you define and measure such a preferred frame?
What do you use for an "absolute" length of a ruler & tick of a clock to establish such measurements?

Now it's your turn to define your preferred absolute frame so everyone can understand why you believe this.

Todd



Offline PaulF

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Really a pleasure to read posts on all sides of the em drive...wish I were in the 0.5% club like many here ;)

Since I tend to go macro...I am visualizing fields whose strength is inversely proportional to spacetime...iow...weak forces can span huge distances. Pehaps we are leaving our preoccupation with strong nuclear forces and heading towards the understanding and utilization of the opposite.  As always, thanks for activating some of my dormant brain cells  8)
Sometimes I think of the QV of space as one giant Quantum Entanglement. Big enough macro?
shell
I'll continue on that note. I See the QV as being a singularity, but a special one. Black holes have their singularity rooted in our spacetime. Now open your mind a little for what i am now proposing. Most will agree QV is something underlying, and that possible extra dimension are probably outside our 4d spacetime, in effect enclosing our spacetime. Imagine the QV as being the 0th dimension( or if you take the other angle, the portal to our mirror anti-universe). The base of our existence. I also would then see a link between the QV and black holes as we know them. Black holes could in this case possibly be the drain for matter and energy in our universe, with the energy flowing back into the QV (assuming QV was responsible for what we now call the big bang/inflation). Thus energy is only being given back as it was given in the first place, and all energy must be given back as the universe reaches maximum entropy. result: no entropy coz thereaint no energy left. Makes me think that although our universe may expand indefinitely, the point at which all energy is depleted from our universe could be the point where there is a new big bang. Endless loop, problem solved as to what came before the big bang. For those who say that expansion within one particular universe could last for infinity even in the case mentioned above, I say no. You can't divide energy or time beyond the planck values. Furthermore X divided by infinity equals zero for all intents and purposes, way beyond planck scales. That division will result in 0, even in our universe. At that point : Big Bang! version 2, or 200, 2 quadrillion. So in fact there are even two mechanisms ensuring an endless Big Bang loop, even if our universe accelerates to oblivion. (probably some holes in that due to CoM or CoE, but please do go nuts on that! :) )
All this popped into my head, just thinking about the relationship between 0 and infinity. These two numbers boggle me more every day. How is it, that our universe which loves 0 and infinity, and in which these two numbers are widely presented and perfectly possible, how is it that these two numbers are the only two we humans with our brilliant minds cannot grasp?

Oh and yes, for some people who may ask the question: Yes, our universe is both expanding spatially and shrinking (in mass and energy) at the same time, if one uses above propositions. Maybe there's even a correllation.
« Last Edit: 06/07/2015 02:47 PM by PaulF »

Offline Rodal

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The ghost of Einstein is going to hunt you down.
...Now, how do you define and measure such a preferred frame? What do you use for an "absolute" length of a ruler & tick of a clock to establish such measurements? Now it's your turn to define your preferred absolute frame so everyone can understand why you believe this. Todd
and hopefully the answer will no longer contain "Einstein" (or other scientists) sprinkled around, intending to add some "gravitas" (pun intended  :) ).  Mentioning the name of Einstein does not really add to the substance of any comment.
From now on, points subtracted for any mention of Einstein, Newton, and other towering figures in an argument  ;) . Points added for using mathematical and physical arguments. :) 
[Examples of added points:
1) deltaMass showing that when acceleration is postulated, velocity has to be obtained from its integral;
2) Todd showing that the photon rocket equation being used represents the force exerted by the radiation pressure of light in free space, which is not the same as the forces and momentum imparted to a massive object].
« Last Edit: 06/07/2015 02:56 PM by Rodal »

Offline wallofwolfstreet

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My reply was in regard to your statement that a non accelerating EMDrive was still a source of Free Energy. Could you please respond to this quote:

Quote
The EMdrive doesn't have to be accelerating at all to act as a free energy machine,

I'll get to responding to your link shortly, which I'm sure you know Shawyer has already made comment on the severe acceleration limits which apply to a superconducting EMDrive.

This is my response to just that quote:

All you need to know for that is the equation Powerout=Thrust*velocity.  If the emdrive goes fast enough, at some point, given it's constant thrust to power ratio, it will be able to produce more energy than it needs to run ==> free energy.  None of Shawyer's constraints on acceleration matter to this issue, because the drive doesn't need to accelerate for the above equation to hold.  If the drive conserves energy as it accelerates from rest up to the breakeven velocity where Pout>Pin, that is all well and good, but it simply isn't enough for CoE, because you don't need acceleration to produce more power than you put in.  It happens completely instantaneously at constant velocity, if it is high enough.

Offline frobnicat

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If EM drives like to be free to accelerate to thrust, so be it, if they don't like to be accelerated to thrust, so be it, either way, a buffering intermediate link can accommodate for those whims, at a very, very, very modest energetic cost. So, a 40s long thrust of 50ÁN for 50W at constant velocity, that is exactly the same conditions as claimed results at Eagleworks, if it is reproducible many times, can be used intermittently to reach break even velocities, in principle (if not in practice). What could possibly make it, in principle, not reproducible many times if it can genuinely work one time ? Growing distance with the lab, or cumulative disturbance of the Force ?
The 40 second long thrust was not at constant velocity. Roughly speaking there was transient (let's say for discussion sake in the remaining of this post, that it was about 2 sec long) with constant velocity and after 2 sec it was (roughly speaking) zero velocity.

So

40s 2 sec long thrust of 0 to 50ÁN linear rise at constant velocity,
   ~40s long thrust of 50ÁN for 50W at zero velocity

You make a distinction between constant velocity and constant zero velocity ? Do you consider frame of lab's ground as a privileged frame, or just a convenient frame ?

Quote

During the 2 second rise, what you see is the damped harmonic oscillator response to a step force

It is not the response of an EM Drive free in space, of course.  What would that be?  We have no idea.  If the EM Drive is an artifact, it won't do anything in space.  If it isn't we have to choose a theory to model it (Shawyer, McCulloch, Notsosureofit, etc.)


What I said is that, whatever is position(t) (wrt the vacuum chamber) in the reported experiments, if it is not the consequence of forces or torques between frustum and ground (vacuum chamber and fixed apparatus, earth), that is if it is the consequence of what I think everybody would agree to call a real linear thrust (a real propulsive force useful in deep space, not just a push on the front window of the spacecraft compensated by my back pushing on the seat, or equivalent) it implies a net total momentum impulse of the drive on pendulum arms, on the flexure bearings, on the fixed apparatus, on the vacuum chamber, on earth globally, and earth globally is a body free falling on an inertial trajectory (neglecting solar wind pressure...). The reported position(t) results of Eagleworks would then (assuming a "real" effect, and trusting proportionality to calibration pulses) imply a net total momentum impulse of roughly 50e-6*40=2e-3 kg.m/s at an energetic cost of roughly 50*40=2000J.

So we already have quite a good idea of "the response of an EM Drive free in space", when the EM Drive is attached to a spacecraft the size and mass of earth. My point is that same position(t) wrt to an inertial frame can be simulated by an appropriate actuator between the EM drive and a much lighter spacecraft (than earth) in free space. The much lighter spacecraft in free space would have a much higher acceleration (and total deltaV during the 40s run) than earth does, but the EM drive assembly couldn't tell the difference from this situation and the one we already know about. Apart from the gravitational field, that is orthogonal to the "axis of operation" (for Eagleworks horizontal setup)... If we want a fully fledged replication of all conditions met in the lab, this is just a little bit more of engineering, use a rotating big enough carrousel to mount the drive at equivalent radial acceleration (everybody seems to agree that equivalence principle still applies).

In the end we have a light spacecraft in deep space, driven by successive 40s pulses of 2e-3 kg.m/s costing 2000J each, minus the energy needed to operate the actuator that makes the EM drive operate in the same apparent kinematic conditions as in the lab. This added cost is that of pushing for 40s a rod at 50ÁN for a distance that correspond to the relative displacement between the accelerating spacecraft and the non accelerating thrusting EM drive. For a 100kg spacecraft (not counting the mass of EM drive) accelerating at 50ÁN/100kg=5e-7m/s▓, upon 40s (while the thrusting EM drive is kept on inertial velocity, synchronised with that of rest of spacecraft at start of process), that gives 0.5*a*t▓=0.4mm. So, ideally, it takes only 50ÁN*0.4mm=20nJ (nano Joules) to simulate the appropriate conditions. After the 40s power on period, the actuator has to put the EM drive back into the original position, please grant me that it's in the same ball park for that to be done in about the same 40s, and total is sufficiently many order of magnitudes below 2000J to accommodate for thermodynamical inefficiencies (+ electronic control...) so that it is not affecting the yield of, roughly, 2e-3 kg.m/s for 2000J, at each successive 40s pulse.

My conclusion is that, apart from the evident interest of ruling out for good any spurious forces with the environment (and validating or refuting the effect as a real linear thrust), we do already have a pretty good idea of how an EM drive like the one used at Eagleworks could work in free space, should it be a real linear thrust effect, and without needing custom theory or interpretation of SR or GR.

And a sequence of 2e-3 kg.m/s pulses for 2000J, even diluted to the extreme (1 month off between 40s runs for instance) could already, like it or not, apparently break CoE, if not in practice, at least in principle (2nd law included).

Quote
...
In a damped oscillator, the force feeds energy into the system. The damping force always takes energy out,
because the damping force always points antiparallel to the velocity.

This energy is negligible when compared to the electrical/RF input. Let's say it is 0   :)

Offline Tetrakis

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To revisit the rocket staging thought experiment, consider two scenarios: EMdrive 1 accelerating itself to its "limiting velocity", say 100 km/s, and a chemical rocket carrying EMdrive 2 accelerating itself to 100 km/s by conventional means. The discussion upthread indicates that if COE is conserved, EMdrive 1, having brought itself up to 100 km/s will not be able to accelerate further. But if the rocket propelling EMdrive 2 "stages" away, releasing its EMdrive payload, would EMdrive 2 be able to accelerate to a higher velocity than EMdrive 1 relative to the "real" inertial reference frame EM1 obeys at their mutual starting point? Where you place the "real reference frame" matters.

This strikes me as a qualitative problem with Traveller's theory. How does EMdrive 1 "know" that if it continues to accelerate, it will violate COE? Your description of some sort of subatomic stress left me totally unconvinced, as it has no empirical backing. If you are going to rely on such a thing, please provide a reference to a peer reviewed journal with an impact factor greater than 1.

Quote from: Traveller
This physical change in the properties of matter alters atomic and sub-atomic spacing,
and energies, identical to how gravity contracts matter falling into a gravity well.
« Last Edit: 06/07/2015 04:19 PM by Tetrakis »

Offline Rodal

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If EM drives like to be free to accelerate to thrust, so be it, if they don't like to be accelerated to thrust, so be it, either way, a buffering intermediate link can accommodate for those whims, at a very, very, very modest energetic cost. So, a 40s long thrust of 50ÁN for 50W at constant velocity, that is exactly the same conditions as claimed results at Eagleworks, if it is reproducible many times, can be used intermittently to reach break even velocities, in principle (if not in practice). What could possibly make it, in principle, not reproducible many times if it can genuinely work one time ? Growing distance with the lab, or cumulative disturbance of the Force ?
The 40 second long thrust was not at constant velocity. Roughly speaking there was transient (let's say for discussion sake in the remaining of this post, that it was about 2 sec long) with constant velocity and after 2 sec it was (roughly speaking) zero velocity.

So

40s 2 sec long thrust of 0 to 50ÁN linear rise at constant velocity,
   ~40s long thrust of 50ÁN for 50W at zero velocity

You make a distinction between constant velocity and constant zero velocity ? Do you consider frame of lab's ground as a privileged frame, or just a convenient frame ?
...
More talk about frames cannot hide the facts that

1) the velocity was not constant in the Eagleworks test from the time that the power was turned on. 

2) same velocity (zero) with the power off for  t<0  , than the velocity with power on for t > 2 s

The velocity changed from zero for t<0 to the (roughly constant non-zero) value during the first 2 sec to the (roughly) zero value in the next ~40 seconds.

Yes, I consider a change in velocity to mean that the velocity was not constant, starting from the time that the power was turned on (t=0 below).



We have, approximately, to first order  (Ahem, not legally speaking  ;) ):

Displacement = rises linearly from t=0 to t=2 sec, and it stays constant x=x0 after that
x = (xo/2 sec)t  0 < t < 2
x = xo                t> 2

Velocity = roughly constant from t=0 to t=2 sec, and it becomes roughly zero after 2 sec until power is turned off
v = (xo/2 sec)    0 < t < 2
v = 0                  t>2                      Change in velocity:  (xo/2 sec) - 0 = (xo/2 sec)



So, this is the history:

Power off  t<0 => velocity = 0 (zero)
Power on  0<t<2 s => velocity =(xo/2 sec) for 2 seconds duration  (it implies deltaV = (xo/2 sec) )
Power on t> 2 s => velocity = 0 (zero)   (it implies deltaV = (xo/2 sec) )

Same velocity (zero) for power off t<0  , than the velocity (zero, again) with power on for t > 2 s
« Last Edit: 06/07/2015 06:10 PM by Rodal »

Offline wallofwolfstreet

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Todd -

I commend you on your efforts to develop a theory that can help to resolve the CoE issue of the emdrive.  I have a comment and a question.

First, a comment:
Quote
The physics is similar in nature to hovering in a Newtonian gravitational field, where Special Relativity
does not apply. The Newtonian gravitational potential Φ has units of (m / s)2
, such that the gradient derivative yields an acceleration vector. It represents the potential energy per unit mass and may be treated identically to the velocity squared in Newtonian kinetic energy, v2 = 2E / m .
   

It appears to me that you have confused the gravitational potential, Φgravity=GM/r, with an actual evaluation of the gravitational potential energy.  The gravitational potential is work done by gravity accelerating a unit mass from infinity to the point of evaluation (http://en.wikipedia.org/wiki/Gravitational_potential).  If you actually wanted the gravitational potential energy, you are missing some mathematical mechanics.  I don't think this changes too much, but it is necessary.

Now for the actual issue:  This theory is a completely testable hypothesis with a trivial experiment.

Take a ball and hold it  above the ground.  It has gravitational potential energy equal to m*g*h.  There is a force between the ball and the earth equal to m*g.  At t =0 release this ball. 

Then:
Thrust to power ratio = mg/(d/dt(mgh))
                                =mg/(mgv),         dh/dt=v
                                =1/v

So at t=0 when we release the ball, the thrust to power ratio for the ball is given by: 1/v = 1/0 = inf.

Therefore, the limiting velocity of the ball is 0!  This ball should not be able to move (by my understanding of your theory), but clearly balls do move.  How can this be resolved?


Offline PaulF

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On a lighter side, should we one day get the bad news that the EMdrive is nothing more than a remodelled microwave oven, we could use that to our advantage. We could build a bad news drive. Bad news always travels faster than light  8)
« Last Edit: 06/07/2015 04:25 PM by PaulF »

Offline PaulF

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OK I must let this out.

-thought experiment-

For all who keep going on about preferred frames:

I believe there is a preferred reference frame, it's just not in our reality. It's outside it. So for all intents and purposes, GR remains intact because it does not take into account extra dimensions in that way. In an earlier post I mentioned I liked to see the QV as a 0-dimensional singularity. What I did not mention was that in that respect I also visualised the QV as being directly linked to at least all our four spacetime dimensions. Let's say we could drill a hole in our spacetime to get to the QV. Because the QV is 0-dimensional you could never tap into a specific region of the QV, it's the same everywhere. the QV simply doesn't have dimensions (as we know it) although we seem to think it has because of our perception of dimensions.

Sorry I forgot to link QV and preferred frame in this explanation. I believe the QV can solely act as frame of reference.
« Last Edit: 06/07/2015 04:50 PM by PaulF »

Offline frobnicat

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If EM drives like to be free to accelerate to thrust, so be it, if they don't like to be accelerated to thrust, so be it, either way, a buffering intermediate link can accommodate for those whims, at a very, very, very modest energetic cost. So, a 40s long thrust of 50ÁN for 50W at constant velocity, that is exactly the same conditions as claimed results at Eagleworks, if it is reproducible many times, can be used intermittently to reach break even velocities, in principle (if not in practice). What could possibly make it, in principle, not reproducible many times if it can genuinely work one time ? Growing distance with the lab, or cumulative disturbance of the Force ?
The 40 second long thrust was not at constant velocity. Roughly speaking there was transient (let's say for discussion sake in the remaining of this post, that it was about 2 sec long) with constant velocity and after 2 sec it was (roughly speaking) zero velocity.

So

40s 2 sec long thrust of 0 to 50ÁN linear rise at constant velocity,
   ~40s long thrust of 50ÁN for 50W at zero velocity

You make a distinction between constant velocity and constant zero velocity ? Do you consider frame of lab's ground as a privileged frame, or just a convenient frame ?
...
More talk about frames cannot hide the fact that the velocity was not constant in the Eagleworks test from the time that the power was turned on. 

The velocity changed from the (roughly constant non-zero) value during the first 2 sec to the (roughly) zero value in the next ~40 seconds.

Yes, I consider a change in velocity to mean that the velocity was not constant, starting from the time that the power was turned on (t=0 below).



We have, approximately, to first order  (Ahem, not legally speaking  ;) ):

Displacement = rises linearly from t=0 to t=2 sec, and it stays constant x=x0 after that
x = (xo/2 sec)t  0 < t < 2
x = xo                t> 2

Velocity = roughly constant from t=0 to t=2 sec, and it becomes roughly zero after 2 sec until power is turned off
v = (xo/2 sec)    0 < t < 2
v = 0                  t>2                      Change in velocity:  (xo/2 sec) - 0 = (xo/2 sec)



So, this is the history:

Power off  t<0 => velocity = 0 (zero)
Power on  0<t<2 s => velocity =(xo/2 sec) for 2 seconds duration  (it implies deltaV = (xo/2 sec) )
Power on t> 2 s => velocity = 0 (zero)   (it implies deltaV = (xo/2 sec) )

Same velocity (zero) for power off t<0  , than the velocity with power on for t > 2 s

All right dr Rodal, this was a preliminary question, probably irrelevant that I shouldn't have made since it distracted us from what I wanted to say. Precisely, the rest of my post made no assumption about constant or non constant velocities. It just says that, given an any measured position(t) displacement of frustum wrt to inertial frame (vacuum chamber, for convenience), provided the integrated consequent momentum exchange through spring stiffness implied a net momentum of given magnitude overall (regardless of rising and falling details, hell, it could widely oscillate all the 40s, at average above 0 wouldn't change the argument) for a given energy. And that this same position(t) displacement of frustum wrt to inertial frame can be artificially recreated to get the same effect (same momentum pulse) to propel a free floating spacecraft, hence nullifying the argument that we don't know how a EM drive could behave in space (if it works at all) short of custom theories. Do you intend to comment on that ? Irrelevant ? Why ?

Offline SeeShells

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Velocity = roughly constant from t=0 to t=2 sec, and it becomes roughly zero after 2 sec until power is turned off
v = (xo/2 sec)    0 < t < 2
v = 0                  t>2                      Change in velocity:  (xo/2 sec) - 0 = (xo/2 sec)



So, this is the history:

Power off  t<0 => velocity = 0 (zero)
Power on  0<t<2 s => velocity =(xo/2 sec) for 2 seconds duration  (it implies deltaV = (xo/2 sec) )
Power on t> 2 s => velocity = 0 (zero)   (it implies deltaV = (xo/2 sec) )

Same velocity (zero) for power off t<0  , than the velocity with power on for t > 2 s

Than I'm kind of confused at what I see here Dr.
http://emdrive.com/dynamictests.html

Offline Rodal

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All right dr Rodal, this was a preliminary question, probably irrelevant that I shouldn't have made since it distracted us from what I wanted to say. Precisely, the rest of my post made no assumption about constant or non constant velocities. It just says that, given an any measured position(t) displacement of frustum wrt to inertial frame (vacuum chamber, for convenience), provided the integrated consequent momentum exchange through spring stiffness implied a net momentum of given magnitude overall (regardless of rising and falling details, hell, it could widely oscillate all the 40s, at average above 0 wouldn't change the argument) for a given energy. And that this same position(t) displacement of frustum wrt to inertial frame can be artificially recreated to get the same effect (same momentum pulse) to propel a free floating spacecraft, hence nullifying the argument that we don't know how a EM drive could behave in space (if it works at all) short of custom theories. Do you intend to comment on that ? Irrelevant ? Why ?

Concerning the argument that we don't know how an EM drive could behave in space, yes there are several possibilities: we don't know whether the experiment is an artifact, and if it is not an artifact there are several competing theories on how it would behave in Space.   The several competing existing theories disagree as to how it would behave in space. I disagree with authors that write that it would result in constant steady acceleration for constant power input, and on that basis extrapolate space trips.  I also disagree with authors that have ad-hoc "theories".   IMHO if the experiments are not an artifact, how it would behave in Space would depend on what enables it to produce thrust, what is the physical process behind it and what is its long-term operation characteristics.
I don't think we can extrapolate now based on a few experiments in partial vacuum conducted at one research center.  If it works, right now is a black box and we have only a few input-output experiments (particularly in partial vacuum) to be able to extrapolate its practical performance.

What can be done (and you have done an excellent job showing this  :) ) is to show what an EM Drive cannot do in Space. (For example: it cannot provide constant acceleration at constant power)
« Last Edit: 06/07/2015 05:53 PM by Rodal »

Offline Blaine

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If anyone likes to mess around with the frustum calculator program here are dimensions you can input that scale very well.  Lets see, here: big diameter - .2314
                                               small diameter - .1180
                                               cavity length - .2286
                                               frequency - 3.3Ghz
                                               Q - 8,000
                                               Po - 26000 watts
« Last Edit: 06/07/2015 05:45 PM by Blaine »
Weird Science!

Offline dustinthewind

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Todd -

I commend you on your efforts to develop a theory that can help to resolve the CoE issue of the emdrive.  I have a comment and a question.

First, a comment:
Quote
The physics is similar in nature to hovering in a Newtonian gravitational field, where Special Relativity
does not apply. The Newtonian gravitational potential Φ has units of (m / s)2
, such that the gradient derivative yields an acceleration vector. It represents the potential energy per unit mass and may be treated identically to the velocity squared in Newtonian kinetic energy, v2 = 2E / m .
   

It appears to me that you have confused the gravitational potential, Φgravity=GM/r, with an actual evaluation of the gravitational potential energy.  The gravitational potential is work done by gravity accelerating a unit mass from infinity to the point of evaluation (http://en.wikipedia.org/wiki/Gravitational_potential).  If you actually wanted the gravitational potential energy, you are missing some mathematical mechanics.  I don't think this changes too much, but it is necessary.

Now for the actual issue:  This theory is a completely testable hypothesis with a trivial experiment.

Take a ball and hold it  above the ground.  It has gravitational potential energy equal to m*g*h.  There is a force between the ball and the earth equal to m*g.  At t =0 release this ball. 

Then:
Thrust to power ratio = mg/(d/dt(mgh))
                                =mg/(mgv),         dh/dt=v
                                =1/v

So at t=0 when we release the ball, the thrust to power ratio for the ball is given by: 1/v = 1/0 = inf.

Therefore, the limiting velocity of the ball is 0!  This ball should not be able to move (by my understanding of your theory), but clearly balls do move.  How can this be resolved?

I guess I have a question myself.  I was wondering what this velocity is with respect to.  Normally when I see a velocity I think that V=V1-V2 where it is a comparison between two frames with different velocity.  My question is if this velocity is a comparison with us and the accelerated frame of space time?  That is we are stationary but the space flowing into the earth is not so we can assign it a relative velocity.  This relative velocity dilates space and time.  Lower to the surface of the earth clocks run slower and space is more contracted because space is flowing faster.  Higher from the surface space is flowing slower with respect to us so clocks are running faster.  So is this velocity a comparison between our frame and the flow of space time's frame?  In that sense in a gravitational field the ball would move?  Further away from gravitational fields where the relative velocity with respect to space time is zero the ball would not accelerate? 

Hmm, be warned.  I think my hypothesis about free space might be flawed.  The reason being if you are falling faster than free space then its resistance would slow you down.  This might make sense for momentum resisting acceleration and assuming free space catches up to you but in the sense of assuming a velocity for in falling space I am not sure it makes sense.  Unless some ones local space moving towards the earth at near light speed could fall faster than the surrounding space.  Do super fluids behave this way? 
« Last Edit: 06/07/2015 06:18 PM by dustinthewind »

Offline SeeShells

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...Than I'm kind of confused at what I see here Dr.
http://emdrive.com/dynamictests.html
Shell, @frobnicat and I were discussing the only EM Drive tests reported having been performed in a partial vacuum.
Correct me if I am wrong but wasn't the test flawed with a leaky cap?

Offline SeeShells

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What can be done (and you have done an excellent job showing this  :) ) is to show what an EM Drive cannot do in Space. (For example: it cannot provide constant acceleration at constant power)
[/quote]
I'd never say cannot as we are not dealing with a black and white situation. It may not provide constant acceleration at constant power, we simply don't know.

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