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.

...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.

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

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 threadhttp://forum.nasaspaceflight.com/index.php?topic=36313.msg1369875#msg1369875I 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.

Quote from: rfmwguy on 06/07/2015 01:54 amReally 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 Sometimes I think of the QV of space as one giant Quantum Entanglement. Big enough macro?shell

Quote from: deltaMass on 06/07/2015 07:19 am...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

...The ghost of Einstein is going to hunt you down.

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:QuoteThe 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.

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

Quote from: frobnicat on 06/07/2015 12:26 am...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 50µN 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 50µN linear rise at constant velocity, ~40s long thrust of 50µN for 50W at zero velocity

...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 50µN 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 ?

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

...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 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.

Quote from: Rodal on 06/07/2015 01:37 amQuote from: frobnicat on 06/07/2015 12:26 am...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 50µN 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 50µN linear rise at constant velocity, ~40s long thrust of 50µN for 50W at zero velocityYou 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 ?...

The physics is similar in nature to hovering in a Newtonian gravitational field, where Special Relativitydoes 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, v_{2} = 2E / m .

Quote from: frobnicat on 06/07/2015 03:54 pmQuote from: Rodal on 06/07/2015 01:37 amQuote from: frobnicat on 06/07/2015 12:26 am...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 50µN 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 50µN linear rise at constant velocity, ~40s long thrust of 50µN for 50W at zero velocityYou 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 thatx = (xo/2 sec)t 0 < t < 2x = xo t> 2Velocity = roughly constant from t=0 to t=2 sec, and it becomes roughly zero after 2 sec until power is turned offv = (xo/2 sec) 0 < t < 2v = 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

Quote from: frobnicat on 06/07/2015 03:54 pmQuote from: Rodal on 06/07/2015 01:37 amQuote from: frobnicat on 06/07/2015 12:26 amVelocity = roughly constant from t=0 to t=2 sec, and it becomes roughly zero after 2 sec until power is turned offv = (xo/2 sec) 0 < t < 2v = 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 sThan I'm kind of confused at what I see here Dr. http://emdrive.com/dynamictests.html

Quote from: Rodal on 06/07/2015 01:37 amQuote from: frobnicat on 06/07/2015 12:26 amVelocity = roughly constant from t=0 to t=2 sec, and it becomes roughly zero after 2 sec until power is turned offv = (xo/2 sec) 0 < t < 2v = 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 sThan I'm kind of confused at what I see here Dr. http://emdrive.com/dynamictests.html

Quote from: frobnicat on 06/07/2015 12:26 amVelocity = roughly constant from t=0 to t=2 sec, and it becomes roughly zero after 2 sec until power is turned offv = (xo/2 sec) 0 < t < 2v = 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 sThan I'm kind of confused at what I see here Dr. http://emdrive.com/dynamictests.html

Velocity = roughly constant from t=0 to t=2 sec, and it becomes roughly zero after 2 sec until power is turned offv = (xo/2 sec) 0 < t < 2v = 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 ?

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:QuoteThe physics is similar in nature to hovering in a Newtonian gravitational field, where Special Relativitydoes 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, v_{2} = 2E / m . It appears to me that you have confused the gravitational potential, Φ_{gravity}=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/vSo 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?

Quote from: SeeShells on 06/07/2015 05:29 pm...Than I'm kind of confused at what I see here Dr. http://emdrive.com/dynamictests.htmlShell, @frobnicat and I were discussing the only EM Drive tests reported having been performed in a partial vacuum.

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