Let me summarize the last couple of pages:Tajmar is reporting a directional force based on the orientation of a piece of electronics in a self contained setup. Might not be in resonance in the can. He also reports similar findings on a MAGA drive that was not predicted to produce a force within the resolving power of his equipment.The Polish researcher found actual downward force (against rising hot air) on a vertical setup. There is additional data that is hard to interpret as data about the horizontal setup has been mixed. I think he is claim about 10 uN horizontal deflection with both the device in a null configuration and a dummy load. It's hard to tell looking at these graphs but I think he is claiming something like a 27 uN horizontal deflection with the device operational.Jamie is claiming a potential 8uN horizontal deflection (what was the power level of this). This seems to be related to the heating of a piece of electronics.TT wants to make sure Jamie is in resonance and is concerned about this motor mode stuff he has been going on about for years.Lots of yelling.So: 1. The EMDrive surrounded by a plastic insulator might not be working.2. Tajmar's student fabricated device might not be in resonance.3. We need to better characterize the approximately 10 uN force that both Jamie and the Polish team are reporting. Let's make sure this we have not detected an anomalous force effect in the wire.

Quote from: ppnl on 06/09/2018 09:19 pmActually if the EmDrive is hovering without pushing against the earth then it is doing work. It is pulling the earth out of its orbit. The acceleration is infinitesimal even by EmDrive standards. But the mass of the earth is huge so the work is substantial.Google gravity tractor.There was something simple I was forgetting about this situation and this is it.It is not the mass of the Earth that makes the work significant though. That is actually irrelevant. Force times distance is the correct equation as TT said, (force times velocity gives power) but since we are moving the Earth, we need to use the reference frame of the sun to get an inertial frame. The Earth is moving at 30km/s so to get best effect, point it straight up at dawn (near the equator, but best latitude is a function of time of year.) 10 mN would be power of 300W. Added to the Earth.

Actually if the EmDrive is hovering without pushing against the earth then it is doing work. It is pulling the earth out of its orbit. The acceleration is infinitesimal even by EmDrive standards. But the mass of the earth is huge so the work is substantial.Google gravity tractor.

Quote from: ppnl on 06/09/2018 09:19 pmActually if the EmDrive is hovering without pushing against the earth then it is doing work. It is pulling the earth out of its orbit. The acceleration is infinitesimal even by EmDrive standards. But the mass of the earth is huge so the work is substantial.Google gravity tractor.Hi ppnl,Correct.If you do the math, the momentum & KE gain of the Earth is very very very small.As our reference frame is the mass the EmDrive is accelerating, the Earth, this equation makes it simple.

Quote from: TheTraveller on 06/10/2018 01:04 amQuote from: ppnl on 06/09/2018 09:19 pmActually if the EmDrive is hovering without pushing against the earth then it is doing work. It is pulling the earth out of its orbit. The acceleration is infinitesimal even by EmDrive standards. But the mass of the earth is huge so the work is substantial.Google gravity tractor.Hi ppnl,Correct.If you do the math, the momentum & KE gain of the Earth is very very very small.As our reference frame is the mass the EmDrive is accelerating, the Earth, this equation makes it simple.No. The Earth is accelerating, so it is not an inertial frame. The simplest and most obvious choice for an inertial frame (or closen enough to one for our needs) is the sun. I did the math. Your equation is simply wrong unless everything is at rest in the frame you choose.Anyway, what happened to "no more theory" from you until you finish your experiment? If you are going to respond to anything like this at the least you could give an actually meaningful answer to how much acceleration is needed for motor mode.

With an EmDrive there is only one frame that is of any interest and that is the frame of the EmDrive.

If we can make any of these gravity/inertia or (G/I) drives work, we can accelerate (or decelerate) our planet's orbital velocity with respect to the sun, given enough time and resources. This capability will then allow humanity to control global climate change "just" by changing the Earth's orbital distance from the sun, thus decreasing or increasing the solar energy it receives from the sun. Neat solution to a lot of climatic problems if doable.Best, Paul M.

Quote from: TheTraveller on 06/10/2018 03:10 pmWith an EmDrive there is only one frame that is of any interest and that is the frame of the EmDrive.Do you realize that this means that it is always "attached" to an inertial reference frame and therefore cannot accelerate? Otherwise its reference frame would be of very little interest (the "easy" equations work only in inertial reference frames).

Quote from: meberbs on 06/10/2018 02:23 pmQuote from: TheTraveller on 06/10/2018 01:04 amQuote from: ppnl on 06/09/2018 09:19 pmActually if the EmDrive is hovering without pushing against the earth then it is doing work. It is pulling the earth out of its orbit. The acceleration is infinitesimal even by EmDrive standards. But the mass of the earth is huge so the work is substantial.Google gravity tractor.Hi ppnl,Correct.If you do the math, the momentum & KE gain of the Earth is very very very small.As our reference frame is the mass the EmDrive is accelerating, the Earth, this equation makes it simple.No. The Earth is accelerating, so it is not an inertial frame. The simplest and most obvious choice for an inertial frame (or closen enough to one for our needs) is the sun. I did the math. Your equation is simply wrong unless everything is at rest in the frame you choose.Anyway, what happened to "no more theory" from you until you finish your experiment? If you are going to respond to anything like this at the least you could give an actually meaningful answer to how much acceleration is needed for motor mode.With an EmDrive there is only one frame that is of any interest and that is the frame of the EmDrive.Mass knows it's inertia. but it does not know it's velocity, momentum nor KE. Those are constructs based on what some other frames sees. The Work, in Joules, needed to be done by a EmDrive to accelerate a Mass for a period of acceleration of t seconds, using a force of N Newtons is given byWork = (N^2 * t^2) / (2 * m)A 100 sec burst of acceleration will require 100x the Joules of Work to be done on the mass as will a 10 sec burst of acceleration. When that 100 sec burst of acceleration stops and some time later another 100 sec burst of acceleration occurs, it will take the same amount of Work to be done as did the 1st 100 sec burst of acceleration. Why? Because mass has no knowledge of it's velocity, a construct that needs another frame of reference. Time to think of the mass as it's own reference frame. BTW if in any other reference frame the dV that occurred during each 100 sec burst of acceleration was recorded, it would have resulted in the same dKE change because the same dV occurred.dKE = (m * dV^2) / 2dp = m * dVTry to understand that here we have a accelerative Force source that accelerate with the mass and there is no mass exhaust, so the accelerative mass stays constant as does the accelerative Force. Plus the accelerated mass does not know it's velocity, only knows it's inertial mass and the property of inertial mass given to it by the universe. Ie it resists being accelerated. Well maybe not always constant Force but that is a story for another day.It Is Time For The EmDrive To Come Out Of The Shadows

dKE = (m * dV^2) / 2

Quote from: meberbs on 06/09/2018 10:58 pmQuote from: ppnl on 06/09/2018 09:19 pmActually if the EmDrive is hovering without pushing against the earth then it is doing work. It is pulling the earth out of its orbit. The acceleration is infinitesimal even by EmDrive standards. But the mass of the earth is huge so the work is substantial.Google gravity tractor.There was something simple I was forgetting about this situation and this is it.It is not the mass of the Earth that makes the work significant though. That is actually irrelevant. Force times distance is the correct equation as TT said, (force times velocity gives power) but since we are moving the Earth, we need to use the reference frame of the sun to get an inertial frame. The Earth is moving at 30km/s so to get best effect, point it straight up at dawn (near the equator, but best latitude is a function of time of year.) 10 mN would be power of 300W. Added to the Earth. MEBERBS:As you've probably already noticed, your previous observation on the EMdrive accelerating Earth leads one to recall Archimedes comment of "Give me the place to stand, and I shall move the earth." https://en.wikiquote.org/wiki/ArchimedesIf we can make any of these gravity/inertia or (G/I) drives work, we can accelerate (or decelerate) our planet's orbital velocity with respect to the sun, given enough time and resources. This capability will then allow humanity to control global climate change "just" by changing the Earth's orbital distance from the sun, thus decreasing or increasing the solar energy it receives from the sun. Neat solution to a lot of climatic problems if doable.Best, Paul M.

Quote from: TheTraveller on 06/10/2018 03:10 pmdKE = (m * dV^2) / 2This equation is wrong. The correct equation is dKe =0.5*m*(v2^2 - v1^2)Your equation happens to get the right answer for only the case that v1 is equal to 0. That other equation you have posted also has that same restriction in addition to assuming a constant force. That equation you derived using the initial rest frame, and it is not applicable to the accelerating frame of the drive like you claim.Kinetic energy is different in every frame. Any equation that tries to calculate an "absolute" value for the work done on the drive is wrong, because by definition the work done is equal to the change in kinetic energy. The v^2 part of the equation keeps the differences from being equal in different frames.There are ways to handle energy when dealing with an accelerating frame, but those are complicated, and your attempts to use them are ignoring all of the complications.

Not interested in frame variant equations that produce a different answer in different frames.

The Work done by an EmDrive accelerating a fixed mass over a time t is always the same value. It is based on the velocity change and not on some arituary initial and final velocity frame varient numbers. It does not vary because some observer in a different frame measures the start and final velocity and then used your frame varient equation to incorrectly calc the KE gain of the accelerated mass.Time to move away from frame varient thinking.The frame invarient equation, Work Joules = (N^2 * t^2) / (2 * m) works perfectly well and does not need to know anything about frame varient start and final velocity.

Quote from: TheTraveller on 06/11/2018 02:40 amNot interested in frame variant equations that produce a different answer in different frames.Velocity, momentum and kinetic energy are all variant between frames. There is no such thing as frame invariant expressions for them by definition.Quote from: TheTraveller on 06/11/2018 02:40 amThe Work done by an EmDrive accelerating a fixed mass over a time t is always the same value. It is based on the velocity change and not on some arituary initial and final velocity frame varient numbers. It does not vary because some observer in a different frame measures the start and final velocity and then used your frame varient equation to incorrectly calc the KE gain of the accelerated mass.Time to move away from frame varient thinking.The frame invarient equation, Work Joules = (N^2 * t^2) / (2 * m) works perfectly well and does not need to know anything about frame varient start and final velocity.No. Your equation obviously doesn't work. A 1 kg object moving at 10m/s has 50 J of kinetic energy. Moving at 20 m/s it has 200 J of kinetic energy. Your equation predicts 50J given a force of 1N and 10 seconds of acceleration. This is obviously different than the difference between these numbers. Since the definition of the work done is the change in energy and only the kinetic energy changed, your equation does not give the work done.

Velocity, momentum and kinetic energy are all variant between frames. There is no such thing as frame invariant expressions for them by definition.