Quote from: SeeShells on 06/07/2015 05:54 pmQuote from: Rodal on 06/07/2015 05:40 pmQuote 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.Correct me if I am wrong but wasn't the test flawed with a leaky cap?what test being flawed? are you referring to the NASA Eagleworks test in partial vacuum ?

Quote from: Rodal on 06/07/2015 05:40 pmQuote 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.Correct me if I am wrong but wasn't the test flawed with a leaky cap?

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

Quote from: SeeShells on 06/07/2015 05:59 pmQuoteWhat 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)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.If it provides constant steady acceleration at constant power, then it can be used to provide free energy, I think that Todd agrees with that too.

QuoteWhat 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)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.

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)

...Somewhere I read or heard someone saying that the capacitors were failing and were even replaced once. No? Did I glean something else. I've read so much lately that it's kind of melding together.

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: TravellerThis 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.

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.

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?

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 .

...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?

Quote from: wallofwolfstreet on 06/07/2015 04:19 pmTodd - 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?You're forgetting the gradient in the thrust-to-power ratio is required to get g. In your example, the power is the energy exchanged between the ball and the gravitational field. If it were stationary at height h, then there is a constant thrust-to-power ratio, but a force must be exerted on the ball to overcome the gradient in that ratio, wrt the center of mass. When you let go of the ball, the gradient is then free to act. Note, this is Newtonian mechanics, not GR. It's explained differently in GR, because the ball is now in geodesic motion, which is in fact described by the derivatives of the metric potentials.

You're forgetting the gradient in the thrust-to-power ratio is required to get g.

If it were stationary at height h, then there is a constant thrust-to-power ratio, but a force must be exerted on the ball to overcome the gradient in that ratio, wrt the center of mass.

When you let go of the ball, the gradient is then free to act.

...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)

Consider the domain of the following to be a field-free region of flat spacetime in which all measured velocities in some arbitrary inertial observer frame are severely subrelativistic. Thus Galilean transformations and Newton's laws suffice for an analysis to first order in this domain. We also consider the motion to be confined to one dimension - i.e. there exist no transverse forces....

...But for an EmDrive in space, there is no road, and there are no tyres. The drive is oblivious to the value 'v' because there is no preferred reference frame, because there is no solid connection to one.Only if a preferred reference frame exists can "F = P/v" be true for an EmDrive in space.But the existence of a preferred reference frame is a direct violation of a core principle of physics - that the physics is the same in all different inertial frames in a flat, field-free spacetime. Therefore, for an EmDrive in space, "F = P/v" can only be true if this core tenet is violated.I rest my case.

...You know, I understand all those words you used, but in that arrangement they make absolutely no sense to me.Also, I'm not unfamiliar with SR and GR, or physics and math in general. That is not where the issue in my understanding lies.QuoteYou're forgetting the gradient in the thrust-to-power ratio is required to get g.What does this mean? What gradient in the thrust-to-power ratio? The thrust-to-power ratio is location-independent, it is a constant throughout space (in fact it isn't even uniquely defined through space, because it is velocity dependent.) It has zero gradient (more accurately, it's gradient is completely undefined. There is no such thing as the gradient in the thrust-to-power ratio). And how does g play into it?QuoteIf it were stationary at height h, then there is a constant thrust-to-power ratio, but a force must be exerted on the ball to overcome the gradient in that ratio, wrt the center of mass.Once again, what does it mean for a force to overcome a gradient in a ratio? This "gradient" has units of s/m^{2} because the ratio of thrust-to-power is in s/m. What equation relates them such that one can "overcome" the other? QuoteWhen you let go of the ball, the gradient is then free to act. The gradient (if we are still talking about the gradient in thrust-to-power) isn't a force. What do you mean by "the gradient is free to act"? Act on what?I know I'm coming off here a little aggressive, but there is nothing for it. The majority of people reading aren't commenting. If they are to understand correctly, I suppose it is up to us commenters to ask the tough questions.

Consider the domain of the following to be a field-free region of flat spacetime...

...I rest my case.

Every particle of matter in both devices still absorbed the increased inertia when it was accelerated by the rocket engine to the new potential, (v - v0)^2. Note, this is the "change" in velocity, relative to where it started from, relative to its rest frame. The rest frame it started in is a preferred frame for that object, but it was not at rest in any "absolute" sense of the word.

Quote from: Tetrakis on 06/07/2015 04:00 pmTo 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.To answer your question, if the rocket engine and the EM Drive have the same thrust-to-power ratio, then switching off the rocket and switching on the EM Drive will make no difference. Just because EM Drive 2 was only along for the ride makes no difference. Every particle of matter in both devices still absorbed the increased inertia when it was accelerated by the rocket engine to the new potential, (v - v0)^2. Note, this is the "change" in velocity, relative to where it started from, relative to its rest frame. The rest frame it started in is a preferred frame for that object, but it was not at rest in any "absolute" sense of the word.

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.

If you read the paper, please understand I'm referring to the gradient of equation (5);(P/F)^2 = (v - v0)^2, Sorry, I see now I was being sloppy. I hadn't finished my first cup of coffee yet this morning. It happens on forums, I'm not as formal with my language as I should be and my thoughts are not always as clearly stated as I think they are at the time. In haste, I may say things incorrectly or abbreviate my thoughts in ways that are not clear to others. Please, on this forum, taken what I say with a grain of salt. If I publish a paper, then that is what I have given thought to and where I attempted to express them clearly.It is not the gradient of F/P, as I mistakenly said above. Equation 5 is correct in the paper we are discussing, refer to that for the Math please.The gradient of (P/F)^2 = v^2 is not a force, it is an acceleration vector that opposes the force when you are accelerating an object, and results in the Newtonian force of gravity when it is the gravitational acceleration;GM/r = (v - v0)^2d/dr GM/r = -GM/r^2 = g m/s^2d/dr (P/F)^2 = -a m/s^2, Where the ratio P/F is only constant at a constant r, but is a variable wrt r. This gradient opposes the thrust. When the force of the thruster acting on the mass is offset by the gradient (-a), the object can no l longer increase its inertial mass, it has reached equilibrium with its own gravity, due to its increased inertial mass. Pushing on the ground at 1g will not move the earth.Todd

Quote from: vulture4If we accept frame invariance (and it is hard not to) then the reactionless thrust cannot vary with velocity. Precisely. That is all I am saying. Todd and Rodal, however, seem to think differently.

If we accept frame invariance (and it is hard not to) then the reactionless thrust cannot vary with velocity.

...The power input P to the device is taken to be a constant and is also known accurately. The device carries its own power supply on board, which delivers constant input power by design.Putting all the above together, it is clear that we are able to directly test the validity of the assertion "F = P/v" from any inertial frame, and that the measured values of F and P will be identical in any and all of these frames. Clearly, however, the measured value of v will be different in different inertial frames....

The power input P to the device is taken to be a constant and is also known accurately.

the measured value of v will be different in different inertial frames

...Now, for a car accelerating along a flat road, we know that F = P/v is a true description of the situation. Given the above, how can that be? The answer is that in the case of the car, there is indeed a preferred frame in respect of the car, and that is the road itself. The car's tyres are in physical contact with the road and thus 'v' has an absolute significance as far as the car's engine is concerned. It is doing work by pushing off against the road and the elementary mechanics of the situation shows that, for constant P, the thrust F will decrease with increasing v, and F = P/v will hold. Equal and opposite momentum is transferred to the road, which of course is not something we notice, given the mass of the Earth....

Quote from: deltaMass on 06/07/2015 08:41 pmQuote from: vulture4If we accept frame invariance (and it is hard not to) then the reactionless thrust cannot vary with velocity. Precisely. That is all I am saying. Todd and Rodal, however, seem to think differently.So what do you think about the momentum argument, i.e. that radiation pressure exerted by a photon depends only on frequency, and not on group velocity?