Author Topic: EM Drive Developments - related to space flight applications - Thread 10  (Read 1114827 times)

Offline PotomacNeuron

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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.

Thank you for summarizing. I have some comments here. I am biased the opposite way as you so it is useful to counter balance with yours.

First, TT suspected there might not be resonance in Tajmar's cavity, probably because there was no thrust. I think the same kind of suspicion  should be cast on the Polish cavity too, because there was also no definite evidence that there was resonance.

Second, Monomorphic's experiment I think was a power on test; there was no microwave involved. 

Third, you said "1. The EMDrive surrounded by a plastic insulator might not be working." This is a strange conclusion, as strange as Shawyer's belief that there must be acceleration for the EmDrive to enter "motor" mode. It is not far from saying that  the EMDrive made by people younger than 50 might not be working. After all, this statement has some support because Shawyer, TT, Paul claimed thrust but Tajmar, the California PhD students and monomorphic didn't.
« Last Edit: 06/10/2018 01:50 PM by PotomacNeuron »
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Offline Star-Drive

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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.
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/Archimedes

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.
Star-Drive

Offline meberbs

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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.

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.

Offline TheTraveller

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.

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 by

Work = (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) / 2
dp = m * dV

Try 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
« Last Edit: 06/10/2018 03:16 PM by TheTraveller »
It Is Time For The EmDrive To Come Out Of The Shadows

Offline wicoe

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With 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).

Offline Bob Woods

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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.
OK, but who gets to drive?  :D

"Shotgun!"
"But Debbie got to drive yesterday."
"Paul, slow down! You're going to hit an asteroid..."
« Last Edit: 06/10/2018 03:28 PM by Bob Woods »

Offline TheTraveller

With 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).

What I meant was mass does not know it's velocity, so making Work calculations based on some other reference frame's observed velocity of the mass is not the reality of the Work done by the EmDrive generated Force to Move a Mass a Distance.

If you wish use the inertial frame what existed just before acceleration started as the initial velocity (Vs), being zero and the inertial frame that is created just as acceleration stops as the final velocity (Vf).

Then KE = (m * (Vf - Vs)^2) / 2 works as does dKE = (m * dV^2) / 2

Or use my equation and avoid using velocity from some external frame. Both methods will generate the same value of Work done and KE Joule gain.

Work = (N^2 * t^2) / 2 * m) then KE Joules = W Joules.

It works and does generate the actual work value that is required to be done to achieve the P-P acceleration of mass.

It Is Time For The EmDrive To Come Out Of The Shadows
« Last Edit: 06/10/2018 03:32 PM by TheTraveller »
It Is Time For The EmDrive To Come Out Of The Shadows

Offline OnlyMe

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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.

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 by

Work = (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) / 2
dp = m * dV

Try 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

TT, the argument above, assumes a completely flat spacetime, that the vacuum or space, or even operation in the earth’s atmosphere, and “empty”... That there is nothing resisting the EmDrive’s acceleration from each inertial frame associated with each drive off/on cycle.

In reality any space we might operate in in the near future is not empty and an object’s velocity relative to the dynamics of the contents of that space, must be considered when attempting to calculate the force required to accelerate an object. When dealing with conventional propulsion technologies this is not as important because the initial force a rocket provides and the lasting inertia of the projectile (satellite or spaceship), is far greater than the resistance the rarefied contents of the vacuum/space is moves through.

And none of this begins to take into consideration the affect of acceleration and/or possibly relativistic velocities through a quantum vacuum with potentially even partially Machian characteristics......

Theoretically all inertial frames are equivalent. Practically all we know is that within our ability to experimentally verify all inertial frames are equal within the limitation of classical velocities. Even our understanding of acceleration outside of particle accelerators is limited to classical limitations.

Offline meberbs

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dKE = (m * dV^2) / 2
This 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.

Offline daveklingler

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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.
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/Archimedes

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.

As a fan of "neat" solutions, I can't help thinking that there are much more elegant, i.e. simpler, ways to solve said problems.  :)

But it remains to be seen whether the assumption that "these gravity/inertia or (G/I) drives" exist is a safe one. For my own part, I would not be so confident yet, and even if a drive can be demonstrated to work consistently, it may be decades before the physics catches up.

I can even imagine a nightmare scenario in which physics acts differently depending on time and/or location because it's dependent on a factor we haven't yet learned to quantify. Perhaps one day a drive will work, and the next day it won't, or it works in Texas but not in New Jersey because the Earth is floating through a clump of something that doesn't interact with normal matter...
« Last Edit: 06/10/2018 11:44 PM by daveklingler »

Offline TheTraveller

dKE = (m * dV^2) / 2
This 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.

It Is Time For The EmDrive To Come Out Of The Shadows
« Last Edit: 06/11/2018 02:42 AM by TheTraveller »
It Is Time For The EmDrive To Come Out Of The Shadows

Offline meberbs

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Not 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.
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.
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.

Offline TheTraveller

Not 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.
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.
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.

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

Mass does not know it's velocity as it is a measurement made vs another frame. The initial (Vs) and final (Vf) velocity of the accelerated mass as measured vs that other frame has no effect on the Work done to accelerate the mass.

Work = (N^2 * t^2) / (2 * m) is frame invarient.
dKE = (m * (Vf - Vs)^2) / 2 is frame invarient.
dp = (m * (Vf - Vs) is frame invarient.

Try this:

60,000kg ship somewhere between Earth and Mars. Has the ability to generate a P-P Force of 60kN. The crew power the drive unit for 100 sec of acceleration.

What is the Work done by the P-P drive accelerating the 60t ships Mass, for 100 seconds, using a Force of 60kN?

Some time later the crew do another 100 sec of acceleration.

What is the Work done by the P-P drive accelerating the 60t ships Mass, for the 2nd 100 seconds, using a Force of 60kN?

BTW the answers have nothing to do with the ship's velocity as measured vs the Earth or Mars or the Sun, etc.

It Is Time For The EmDrive To Come Out Of The Shadows
« Last Edit: 06/11/2018 03:30 PM by TheTraveller »
It Is Time For The EmDrive To Come Out Of The Shadows

Offline Chris Bergin

Time for a new thread.

https://forum.nasaspaceflight.com/index.php?topic=45824.0

Traveller, no more of your spammy "Time to come out of the shadows" in every post. It's not a cult (well, it probably is with this EM Drive stuff :D) Change your signature to that if you want, then it's less in people's faces.


NOTE. I WILL BE KEEPING THIS THREAD UNLOCKED FOR A FEW DAYS TO ALLOW FOR POSTS IN THIS THREAD TO BE QUOTED IN THE NEW THREAD, BUT NO MORE POSTS IN THIS THREAD OR THEY'LL BE REMOVED! :o
« Last Edit: 06/11/2018 05:51 PM by Chris Bergin »

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