Author Topic: Woodward's effect  (Read 482003 times)

Offline GeneralRulofDumb

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Re: Woodward's effect
« Reply #1460 on: 07/19/2018 12:38 PM »
Quote
The Harry Bull Reaction Motor is not really a slip-stick effect (asymmetrical frictional resistance) like the classic Dean Drive. Instead, it relies upon one of the collisions being elastic and the other inelastic since there is a damper on one side and not on the other.

I ran some simulations a while back to see how something like this might produce a thrust signature on a torsional pendulum. I also built an "asymmetric shaker" device using a voice coil actuator and a ~50g stainless steel reaction mass that generated ~2uN at ~10KHz. Since then I've built a device that I hope will generate ~10uN+ and also improved the simulation. You can see that here:

I guess if this would work, we would be flying through space on washing machines set to a high spin cycle.
Or on jack hammers.
Devices like these only (seem to) produce 'thrust' when fixed to a solid base, i.e. through friction or springs.
However there's nothing in outer space to affix them to.

MET seems to use FOAM as a means to 'propel against'.
Accelerate a vessel in outer space with a MET on board in one direction, and all the matter in the universe (connected through inertia) accelerates in the other direction.
A bit like accelerating your car accelerates the earth (or a piece of a tectonic plate) in the opposite direction, immeasurably as it may be.
This also would mean no violation of over unity energy creation.

Anyway, it might be that pursuing this 'Harry Bull Reaction Motor' has its merits and therefore might warrant creating its own discussion group.

Offline ThinkerX

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Re: Woodward's effect
« Reply #1461 on: 07/19/2018 11:59 PM »
Quote
I guess if this would work, we would be flying through space on washing machines set to a high spin cycle.
Or on jack hammers.
Devices like these only (seem to) produce 'thrust' when fixed to a solid base, i.e. through friction or springs.
However there's nothing in outer space to affix them to.

MET seems to use FOAM as a means to 'propel against'.
Accelerate a vessel in outer space with a MET on board in one direction, and all the matter in the universe (connected through inertia) accelerates in the other direction.
A bit like accelerating your car accelerates the earth (or a piece of a tectonic plate) in the opposite direction, immeasurably as it may be.
This also would mean no violation of over unity energy creation.

Anyway, it might be that pursuing this 'Harry Bull Reaction Motor' has its merits and therefore might warrant creating its own discussion group.

The closest valid real world comparison I can make is using planetary flyby's to increase a spacecraft's velocity or alter course - something NASA seems to have a knack for.  That process, like the one claimed here, is essentially a transfer of momentum (kinetic energy) - huge for the spacecraft, effectively negligible for the planet. 

But, I keep seeing issues arising from especially powerful local gravitational sources for this mechanism.  (aka, the moon)
 




Offline GeneralRulofDumb

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Re: Woodward's effect
« Reply #1462 on: 07/20/2018 08:15 AM »
The closest valid real world comparison I can make is using planetary flyby's to increase a spacecraft's velocity or alter course - something NASA seems to have a knack for.  That process, like the one claimed here, is essentially a transfer of momentum (kinetic energy) - huge for the spacecraft, effectively negligible for the planet. 

But, I keep seeing issues arising from especially powerful local gravitational sources for this mechanism.  (aka, the moon)

I don't think that's a valid or even useful comparison.
Slingshotting using a planets' (or moons') gravity well relies on gravity.
The MET relies on inertia for it to work.
Gravity and inertia are often mentioned synonymously, but they are definitely not the same.
Although they both are directly related to a bodies mass, they constitute two very different phenomenons.

The real question is: what is inertia end why do bodies with mass exhibit it ?
Is all matter in the universe connected through inertia ?
If so, we should be able to use a MET to 'grab onto' all matter and create propellantless thrust.

Offline ppnl

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Re: Woodward's effect
« Reply #1463 on: 07/21/2018 03:57 AM »
Quote
The Harry Bull Reaction Motor is not really a slip-stick effect (asymmetrical frictional resistance) like the classic Dean Drive. Instead, it relies upon one of the collisions being elastic and the other inelastic since there is a damper on one side and not on the other.

I ran some simulations a while back to see how something like this might produce a thrust signature on a torsional pendulum. I also built an "asymmetric shaker" device using a voice coil actuator and a ~50g stainless steel reaction mass that generated ~2uN at ~10KHz. Since then I've built a device that I hope will generate ~10uN+ and also improved the simulation. You can see that here:

I guess if this would work, we would be flying through space on washing machines set to a high spin cycle.
Or on jack hammers.
Devices like these only (seem to) produce 'thrust' when fixed to a solid base, i.e. through friction or springs.
However there's nothing in outer space to affix them to.

MET seems to use FOAM as a means to 'propel against'.
Accelerate a vessel in outer space with a MET on board in one direction, and all the matter in the universe (connected through inertia) accelerates in the other direction.
A bit like accelerating your car accelerates the earth (or a piece of a tectonic plate) in the opposite direction, immeasurably as it may be.
This also would mean no violation of over unity energy creation.

Anyway, it might be that pursuing this 'Harry Bull Reaction Motor' has its merits and therefore might warrant creating its own discussion group.

No, the MET still has several problems here that lead to local power generation.

Imagine an electric car traveling at some speed. The amount of energy needed to accelerate more depends on how fast the car is already going. Thus you can measure how fast the car is going by feeding a measured amount of energy into the motor and measuring the acceleration force it causes. Or you could tap the regenerative braking and measure the amount of energy created. This does not violate conservation of energy because you are just tapping into the difference in velocity between the earth and the car. It creates the illusion of a preferred frame. But that frame is not universal.

If the MET is reacting against something then it should be able to do the same thing. You should be able to generate a huge amount of energy locally that you can use any way you choose. You can claim that it does not violate conservation of energy but it is still effectively free infinite energy. You have also created a preferred frame that is the same for all. Also if your velocity with respect to the preferred frame is high then it can take a huge amount of energy to get trivial amounts of acceleration. Or by going in the other direction you can get huge amounts of acceleration with huge amounts of free energy. Just like a braking electric car.

You can attempt to get around this by saying that the MET acts the same in all frames of reference. That means, unlike the car, constant power means constant acceleration. This gives a violation of conservation of energy that cannot be easily fixed by reference to something it is pushing against. If you are pushing against something then how fast you are moving matters. If it does not matter then you break physics.

You could fix all of this by just saying the MET reacts against only local masses. For example a MET on the surface of the earth could accelerate upwards by pushing against the earth with some hypothetical inertial field. This way you could conserve momentum and energy and still not have any preferred frames.  It is pure science fiction and the experimental evidence for anything MET like is so weak as to be properly ignored. But at least this is consistent with the experimental evidence such as it is. And it does not immediately break all of physics or create an infinite energy supply. 


Offline Bob Woods

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Re: Woodward's effect
« Reply #1464 on: 07/21/2018 04:42 AM »
Imagine an electric car traveling at some speed. The amount of energy needed to accelerate more depends on how fast the car is already going.
Only at relativistic speeds where mass is increasing, otherwise Newton rules. F=ma.


This is why the MEGA drive expects to be limited to a small, but still significant fraction of c.

Offline meberbs

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Re: Woodward's effect
« Reply #1465 on: 07/21/2018 04:54 AM »
Imagine an electric car traveling at some speed. The amount of energy needed to accelerate more depends on how fast the car is already going.
Only at relativistic speeds where mass is increasing, otherwise Newton rules. F=ma.


This is why the MEGA drive expects to be limited to a small, but still significant fraction of c.
No, ppnl's statement is true at all speeds. In the non-relativistic situation, kinetic energy is 0.5*m*v^2 which is nonlinear with velocity.

Offline Bob Woods

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Re: Woodward's effect
« Reply #1466 on: 07/21/2018 05:04 AM »
Imagine an electric car traveling at some speed. The amount of energy needed to accelerate more depends on how fast the car is already going.
Only at relativistic speeds where mass is increasing, otherwise Newton rules. F=ma.


This is why the MEGA drive expects to be limited to a small, but still significant fraction of c.
No, ppnl's statement is true at all speeds. In the non-relativistic situation, kinetic energy is 0.5*m*v^2 which is nonlinear with velocity.
My bad. Should not post after beer.

Offline soms42

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Re: Woodward's effect
« Reply #1467 on: 07/21/2018 01:59 PM »
"This is why the MEGA drive expects to be limited to a small, but still significant fraction of c."

This is weird. Speed is relative to an observer, so it can never be limited (except by c).

edit:
Let me expain the a little bit. When a mach drive is accelerating a space ship, the space ship keeps on accelerating getting relativistic velocities relative to the non-accelerated observer.
There is no magical speed limit, except for the available energy on the spaceship needed to power the mach-drive.
« Last Edit: 07/21/2018 02:15 PM by soms42 »

Offline ppnl

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Re: Woodward's effect
« Reply #1468 on: 07/22/2018 02:17 AM »
"This is why the MEGA drive expects to be limited to a small, but still significant fraction of c."

This is weird. Speed is relative to an observer, so it can never be limited (except by c).

edit:
Let me expain the a little bit. When a mach drive is accelerating a space ship, the space ship keeps on accelerating getting relativistic velocities relative to the non-accelerated observer.
There is no magical speed limit, except for the available energy on the spaceship needed to power the mach-drive.

Yes that is true. But remember the energy you need goes up with the square of the velocity. So getting up to even a small fraction of the speed of light is going to be very very hard.

And you still have that problem with the prefered frame of reference.

Offline flux_capacitor

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Re: Woodward's effect
« Reply #1469 on: 07/23/2018 05:57 PM »
Physicist Mike McCulloch thinks that the "Woodward effect" showing an anomalous thrust with Mach Effect Thrusters (METs) is actually an effect of quantised inertia, relying on the dampening of Unruh radiation, making these thrusters "horizon drives":

• Mike McCulloch (25 June 2018): "Does QI Predict The Woodward Effect?"

Offline soms42

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Re: Woodward's effect
« Reply #1470 on: 07/23/2018 07:16 PM »
"This is why the MEGA drive expects to be limited to a small, but still significant fraction of c."

This is weird. Speed is relative to an observer, so it can never be limited (except by c).

edit:
Let me expain the a little bit. When a mach drive is accelerating a space ship, the space ship keeps on accelerating getting relativistic velocities relative to the non-accelerated observer.
There is no magical speed limit, except for the available energy on the spaceship needed to power the mach-drive.

Yes that is true. But remember the energy you need goes up with the square of the velocity. So getting up to even a small fraction of the speed of light is going to be very very hard.

And you still have that problem with the prefered frame of reference.

The applied energy is in the reference frame of the accelerated object, so E=.5mv² does not apply this way.

Offline meberbs

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Re: Woodward's effect
« Reply #1471 on: 07/23/2018 10:45 PM »
"This is why the MEGA drive expects to be limited to a small, but still significant fraction of c."

This is weird. Speed is relative to an observer, so it can never be limited (except by c).

edit:
Let me expain the a little bit. When a mach drive is accelerating a space ship, the space ship keeps on accelerating getting relativistic velocities relative to the non-accelerated observer.
There is no magical speed limit, except for the available energy on the spaceship needed to power the mach-drive.

Yes that is true. But remember the energy you need goes up with the square of the velocity. So getting up to even a small fraction of the speed of light is going to be very very hard.

And you still have that problem with the prefered frame of reference.

The applied energy is in the reference frame of the accelerated object, so E=.5mv² does not apply this way.
The energy is applied, period. The energy coming from the energy source is not frame dependent in general so specifying a frame changes nothing.

The "frame of the accelerated object" is by definition an accelerating frame. An accelerating frame is not an inertial reference frame and simply shouldn't be used unless you want a headache from the corrections you need to apply. Instead, it is easier to just pick an inertial frame to calculate from and 0.5*m*v^2 applies.

Offline ThinkerX

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Re: Woodward's effect
« Reply #1472 on: 07/24/2018 12:39 AM »
Quote
The "frame of the accelerated object" is by definition an accelerating frame. An accelerating frame is not an inertial reference frame and simply shouldn't be used unless you want a headache from the corrections you need to apply. Instead, it is easier to just pick an inertial frame to calculate from and 0.5*m*v^2 applies.

Ok, my brain just melted here trying to grasp all this.  (Then again, long annoying day at work)

Could Woodward's device very roughly be considered the equivalent of a sail on an old fashioned sailboat?  Doing little else but capturing inertial force?

Offline Augmentor

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Re: Woodward's effect
« Reply #1473 on: 07/24/2018 04:20 AM »
Quote
The "frame of the accelerated object" is by definition an accelerating frame. An accelerating frame is not an inertial reference frame and simply shouldn't be used unless you want a headache from the corrections you need to apply. Instead, it is easier to just pick an inertial frame to calculate from and 0.5*m*v^2 applies.

Ok, my brain just melted here trying to grasp all this.  (Then again, long annoying day at work)

Could Woodward's device very roughly be considered the equivalent of a sail on an old fashioned sailboat?  Doing little else but capturing inertial force?

Perhaps after a few beers or a bottle of wine.

A sailboat captures momentum from external sources.  Mach effects requires separation from the Machian universe for a transient moment. So Mach effect craft are more like an electric motor boat.

The brilliance of Woodward's device? The MEGA captures momentum from internal sources in part by disconnecting from external sources.

The method of doing is a calculation within General Relativity, not a new theory.

Woodward's MEGA (formerly MET) is a resonant closed system where mechanical acceleration is achieved at the molecular level in a dielectric, typically PZT. During resonance, a second source of energy, electricity applied to a capacitor (PZT) makes a transient change in the oscillating mass, an acoustic particle-wave,  in a rather short amount of time. The transient momentum is transferred to a bulk mass at one end, and the mass is then recycled with a return trip.

So there is a external acceleration within the universe, and in addition, an internal change resulting in transient  mass and therefore momentum. Normally, we would say this is Newtonian, perhaps non-Newtonian, yet at this point, not relativistic.

The relativistic change is due to the local separation from the Machian universe and occurs at the molecular and atomic level aka the mesoscopic level of physics where both macro effects are produced by micro effects. Charge is invariant (does not vary in direction or magnitude). Bonds get stretched (van der Waals, covalent), magnetic field lines disconnect and reconnect, and atoms get energized.

For now, begin with the resonance of the capacitor can be described as a directionally forced, damped harmonic oscillator commonly found in micro and nano mechanical resonators.

m (x'' + (2 pi f(0)/Q)  x' + 2pi f(0)^2  x) = F cos (f t)

Then one needs to simply add transient mass in one direction, calculate the change in momentum to the bulk mass at one end.

A more meaningful equation addresses the non-relativistic kinematics with an equation of motion. For relativistic kinematic solution, please see the NASA NIAC report where Dr. Fearn did the math for coupled oscillators.

The result is that for a kinematically accelerating system (PZT capacitor), electrically induced internal changes within the PZT cell produce a local acceleration. While the macro system is accelerating, an energy pulse produces an internal acceleration, a nested forces calculation that provides for what Woodward calculates in STARGATES and the MSAS book.

F' is the change in force  d^2(mv)/dt^2 = (d/dt (d(m v)/dt) = d/dt (m a) = F'

Woodward derives the following in his book, and in the STARGATES paper.
For nested systems especially where relativistic effects at the micro level occur,
then one has to calculate the induced change in force resulting in:

F'  = A a^2 + B j v

v is velocity
a is acceleration
j is jerk/jolt == a change in acceleration
A and B are constants to highlight kinematics. However, they could be functions too.

The MEGA is a motor boat that slowly accelerates, for now. The fun part is
the MEGA accelerates not just for seconds or minutes like a chemical rocket
nor for weeks or months like a Lorentz electric rocket, but for accelerates
for years and decades.

Fueled by fusion, the MEGA motor boat needs more power. The current path
is to amplify the power signal input per unit, increase the efficiency of the GI
output, and scale by arrays with multiple units to useful thrust and control
levels.

Jump Rope Rhyme

Motorboat, motorboat, go so slow
Motorboat, motorboat, go so fast
Motorboat, motorboat step on the gas!
VROOOOOOM!

Sailships? I like tall ships. Project Starshot? It may work if the
lasers and masers don't turn the craft into a burnt s'more.

David



Offline soms42

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Re: Woodward's effect
« Reply #1474 on: 07/24/2018 06:13 PM »
"This is why the MEGA drive expects to be limited to a small, but still significant fraction of c."

This is weird. Speed is relative to an observer, so it can never be limited (except by c).

edit:
Let me expain the a little bit. When a mach drive is accelerating a space ship, the space ship keeps on accelerating getting relativistic velocities relative to the non-accelerated observer.
There is no magical speed limit, except for the available energy on the spaceship needed to power the mach-drive.

Yes that is true. But remember the energy you need goes up with the square of the velocity. So getting up to even a small fraction of the speed of light is going to be very very hard.

And you still have that problem with the prefered frame of reference.

The applied energy is in the reference frame of the accelerated object, so E=.5mv² does not apply this way.
The energy is applied, period. The energy coming from the energy source is not frame dependent in general so specifying a frame changes nothing.

The "frame of the accelerated object" is by definition an accelerating frame. An accelerating frame is not an inertial reference frame and simply shouldn't be used unless you want a headache from the corrections you need to apply. Instead, it is easier to just pick an inertial frame to calculate from and 0.5*m*v^2 applies.

I am starting to belief this is a confusion of tongues.
Of course, for the non accelerated observer the energy of the accelerated spaceship increases with  0.5*m*v^2 . However, the applied power in the spaceship is constant.

Offline Jim Davis

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Re: Woodward's effect
« Reply #1475 on: 07/24/2018 06:46 PM »
Of course, for the non accelerated observer the energy of the accelerated spaceship increases with  0.5*m*v^2 . However, the applied power in the spaceship is constant.

The problem is that different observers see different increases in the kinetic energy of the accelerated spaceship. So if the applied power in the spaceship is constant, energy is not being conserved in all reference frames. Indeed, it is only being conserved in one reference frame.

That is a problem.

Offline ThinkerX

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Re: Woodward's effect
« Reply #1476 on: 07/24/2018 10:34 PM »
Quote
The problem is that different observers see different increases in the kinetic energy of the accelerated spaceship. So if the applied power in the spaceship is constant, energy is not being conserved in all reference frames. Indeed, it is only being conserved in one reference frame.

That is a problem.

Seems to me you'd run into that frame problem with other, less controversial drives - like, say, a Buzzard Ramjet.  Say you had a workable such ramjet (yes, whole can of space worms, there, but...) and say said ramjets controls have just two speeds: 'stop' and 'full,' with all fuel external to the craft.  From a frame perspective, wouldn't you have the same issues as with the Mach Drive? 

With the frame issue, its almost like we're getting into an area where even ordinary rockets can't work.

Offline meberbs

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Re: Woodward's effect
« Reply #1477 on: 07/24/2018 11:01 PM »
Quote
The problem is that different observers see different increases in the kinetic energy of the accelerated spaceship. So if the applied power in the spaceship is constant, energy is not being conserved in all reference frames. Indeed, it is only being conserved in one reference frame.

That is a problem.

Seems to me you'd run into that frame problem with other, less controversial drives - like, say, a Buzzard Ramjet.  Say you had a workable such ramjet (yes, whole can of space worms, there, but...) and say said ramjets controls have just two speeds: 'stop' and 'full,' with all fuel external to the craft.  From a frame perspective, wouldn't you have the same issues as with the Mach Drive? 

With the frame issue, its almost like we're getting into an area where even ordinary rockets can't work.
The fuel is a medium that has its own kinetic energy in the Bussard ramjet. The energy of the fuel is frame dependent. To produce a constant force, the ramjet would need to output a variable amount of energy depending on its speed relative to the medium. All frames agree on the amount of energy output needed to produce a given force, because with the medium moving separately from the craft, the terms cancel out.

Might be more clear if you pick some numbers and work out an example. With the change in speed of the propellant and the ship being in different directions, the kinetic energy changes balances to give a consistent result for all frames.

Offline ppnl

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Re: Woodward's effect
« Reply #1478 on: 07/24/2018 11:07 PM »
Quote
The problem is that different observers see different increases in the kinetic energy of the accelerated spaceship. So if the applied power in the spaceship is constant, energy is not being conserved in all reference frames. Indeed, it is only being conserved in one reference frame.

That is a problem.

Seems to me you'd run into that frame problem with other, less controversial drives - like, say, a Buzzard Ramjet.  Say you had a workable such ramjet (yes, whole can of space worms, there, but...) and say said ramjets controls have just two speeds: 'stop' and 'full,' with all fuel external to the craft.  From a frame perspective, wouldn't you have the same issues as with the Mach Drive? 

With the frame issue, its almost like we're getting into an area where even ordinary rockets can't work.

Ordinary rockets carry their reaction mass with them. That means in the early stage of a rocket accelerating most of the energy is pumped into the kinetic energy of the unused fuel. To reach orbit like 90% of your weight is fuel. To go faster the amount of fuel quickly increasses to the point that it becomes impractical. This is called tyranny of the rocket equation. At no point does a rocket violate conservation of momentum or energy from the point of view of any frame of reference.

A  Buzzard Ramjet is a little different but again it never violates conservation of momentum or energy from the point of view of any frame of reference.


Offline Jim Davis

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Re: Woodward's effect
« Reply #1479 on: 07/25/2018 06:39 PM »
With the frame issue, its almost like we're getting into an area where even ordinary rockets can't work.

Oh, no. Rockets (and Bussard ramjets and all other know propulsion devices) conserve energy and momentum in all frames. I'll do the math for a low speed rocket to demonstrate but the conclusions hold for other devices at all speeds.

The control volume encloses the ship and its exhaust at all times. Nothing crosses the boundaries of the control volume.

The conservation of mass equation is given by

M_I = m + m_e

where

M_I = initial mass of ship
m = mass of ship after time, t
m_e = mass of exhaust after time, t 

Differentiating

0 = dm + dm_e (Equation 1)

The conservation of momentum is given by

M_I V_I = m v + Integral[(v - v_e) dm_e]

where

V_I = initial velocity of ship
v = velocity of ship after time, t
v_e = velocity of exhaust with respect to ship

Differentiating

0 = m dv + v dm + (v - v_e) dm_e (Equation 2)

Substitute Equation 1 into Equation 2

0 = m dv + v dm - v dm + v_e dm

which reduces to

0 = m dv + v_e dm

or

dv = - v_e dm / m  (Equation 3)

which can be integrated to give the familiar low speed rocket equation.

The energy equation is given as

E_I + M_I (V_I)^2 / 2 = e + m v^2 / 2 + Integral[(v - v_e)^2 / 2 dm_e]

where

E_I = initial non-kinetic energy of ship
e = non-kinetic energy of ship after time, t

Differentiating

0 = de + m v dv + v^2 / 2 dm + (v - v_e)^2 / 2 dm_e

expanding terms

0 = de + m v dv + v^2 / 2 dm + v^2 / 2 dm_e + (v_e)^2 / 2 dm_e - v v_e dm_e

Substitute Equation 1 into the above

0 = de + m v dv + v^2 / 2 dm - v^2 / 2 dm - (v_e)^2 / 2 dm + v v_e dm

which reduces to

0 = de + m v dv - (v_e)^2 / 2 dm + v v_e dm

Substitute Equation 3 into the above

0 = de - v v_e dm - (v_e)^2 / 2 dm + v v_e dm

which reduces to

de = (v_e)^2 / 2 dm

Differentiating with respect to time, t, gives

P = de / dt = (v_e)^2 / 2 dm / dt

where

P = power consumption of the ship

So for a rocket power consumption depends only on v_e, which all frames of reference agree on, and mass consumption dm / dt, again which all frames of reference agree on. Energy is conserved in all frames.

Note the importance of the dm_e terms in the above which allow all conservation laws to be obeyed.

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