Author Topic: Woodward's effect  (Read 803011 times)

Offline dustinthewind

  • Full Member
  • ****
  • Posts: 902
  • U.S. of A.
  • Liked: 313
  • Likes Given: 355
Re: Woodward's effect
« Reply #500 on: 09/24/2015 04:50 pm »
just curious have you seen the NBF article:

http://nextbigfuture.com/2015/08/theory-explaining-electromagnetic.html

it seems topical to your post.

also in away topical to Woodward's book where he is talking about electron modeling and the problems of eliminating divergences or infinities.

I would imagine that is how radiation is contained in the cavity.  Currents in the cavity cancel the radiation from the injecting antenna.  Notice in the Anapole the currents in the toroid flow in the same direction as the outer torrid.  He mentions they are out of phase.  They are also separated by space so you have the space time relation ship of delayed information.  I would think the radiation emitted by each should have opposite electric polarization but overlapping.  You can imagine both types of radiation escaping through the cavity but effectively canceling each other out as they travel.  The distances of the currents in the anapole don't look exactly the same and symmetric but maybe it works for far field cancellation or the drawing is a bit off from the paper.  In our case with the cavity we are using transverse electric field where as the type they are using is transverse magnetic as you can tell from the circular magnetic field inside the toroid.  This configuration would lead to charge separation inside the cavity. 

Maybe the nucleus reacts in such a way as to emit radiation that cancels light emitted from the electron but I couldn't say for sure. 

I would think it more of a possibility for the atom that the electron is like a string that unravels it self around the atom.  The electron then encompassing the atom as a rotating symmetric string should then not emit radiation?  Just a guess on my part. 
« Last Edit: 09/24/2015 08:16 pm by dustinthewind »
Follow the science? What is science with out the truth.  If there is no truth in it it is not science.  Truth is found by open discussion and rehashing facts not those that moderate it to fit their agenda.  In the end the truth speaks for itself.  Beware the strong delusion and lies mentioned in 2ndThesalonians2:11.  The last stage of Babylon is transhumanism.  Clay mingled with iron (flesh mingled with machine).  MK ultra out of control.  Consider bill gates patent 202060606 (666), that hacks the humans to make their brains crunch C R Y P T O. Are humans hackable animals or are they protected like when Jesus cast out the legion?

Offline aceshigh

  • Full Member
  • ****
  • Posts: 792
  • Liked: 269
  • Likes Given: 22
Re: Woodward's effect
« Reply #501 on: 09/24/2015 10:30 pm »
Dustinthewind, maybe you should try to join Woodward's mailing list?

I did my thesis on the nature of the magnetic field and we were trying to detect if it could rotate axially.  The idea was to take a solenoid and rotate it axially and imagine the magnetic field lines would rotate axially with it.   As a result by (v x B) of these rotating magnetic field lines an electric field should come off the solenoid with a 1/r^2 dependence.  We set up a capacitor and estimated what we should see.  The experiment was done with a permanent magnet and solenoid.  Both resulted in a negative indication of magnetic field line rotation.  After some investigation it became more obvious that the magnetic field shouldn't rotate but rather radiates to change orientation.  The nature of information is to radiate in a radial direction and if you look into the electric field of a charge moving in a circle, you will notice the electric field doesn't rotate with the charge.  The magnetic field being what describes the relativistic behavior of the electric field (Edward Purcell Electricity and Magnetism: electric field pancaking) it becomes obvious that the magnetic field of a charge also can't rotate but rather also radiates to change orientation. 

After this it became evident the speed limits in changes of the magnetic field (time retarded information) and that electromagnetic propulsion should be possible.  I began with a diagram of current in two wires and mapped out the time retarded behavior of two wires a distance of 1/4 lambda which had their currents 90 degrees out of phase and got a unidirectional force.  Later, it was realized a current flows because of charge separation and current flow induces charge separation so this was later included in the wire diagram.  A unidirectional force from the static charge was also found but appears to oppose the magnetic unidirectional force on the wires.  It turns out this dual wire diagram is a phased array antenna and indeed you do get propulsion even with the opposing forces (magnetic and static electric) which is photon propulsion.  Phased array antennas can direct radiation in desired directions just by modifying the phase of currents in straight antenna.  I latter stumbled across a patent, the EM drive and the Woodward effect which I highly suspect is connected to what I am dealing with.  No idea what is behind the EM drive but maybe they are connected. 

After watching the EM drive for a while it became evident of a way to cancel the opposing static electric force using resonating cavities.  That is if I take the antenna of the phased array antenna and loop it then for a standing wave I cancel the charge separation and only the magnetic force works.  This happens inside of a resonant cavity for transverse electric fields, one mode being TE011 for a cylindrical cavity.  Instead of energy alternating in the cavity between current and charge separation (magnetic and capacitance) you get energy alternating between current and light stored in the cavity (J and -dB/dt=light=curl E).  To excite this mode I suspect we would need an antenna inside the cavity shaped in the shape of the mode we want to excite.  That is the wire should be in the shape of the mode induced and 1/4 lambda from the inside plate for constructive interference of reflected energy.  I don't consider myself an expert in microwave engineering so remember that.  I am using my intuition but I know the electric field will be oriented in the direction of the wire and that will induce the currents in the cavity. 

The other issue is matching the frequency of two adjacent cavities.  For TE modes with no current flow from the flat circle plate to the side walls we don't have to worry about electrical connectivity of one of the end plates.  Changing the distance of an end plate changes the frequency in the cavity.  We can monitor the frequency in one cavity and match it with the other cavity.  We can then control the phase to the current by the phase of the injected radiation and also by increasing or decreasing the frequency of one cavity by moving the cavity plate then moving the plate back and leaving the light slightly out of phase  (probably easier to control the phase of the injected radiation).  Current in one cavity perceives the current in the lower cavity as repulsive (magnetic).  The cavity plates are separated by a distance of 1/4 lambda (wavelength).  Due to information delay of apparent current and the phase relation the current in the other cavity perceives its partner to be attractive.  So we have an asymmetry in force over complete cycles (see paper 1 for diagrams).  Essentially we are playing with time and space and taking advantage that the information can not travel beyond a speed limit of light.  Considering only the magnetic force is now pulling, we may get forces beyond photon propulsion and we may find out there is a connection to space time manipulation or gravitational in nature.  What is interesting about the dual cavity experiment proposed is that there might be no radiation (dynamic magnetic field -dB/dt) emitted as the radiation should be trapped inside the cavities. At the same time there isn't really a reason why the magnetic force between the cavities shouldn't be there as they should still observe the relativistic dipole distribution of current associated with the static magnetic field.  This static magnetic field should be observable outside a resonant cavity just by holding up a magnetic field sensor to the base.  A voltage should be observed to osculate at the frequency of the current in the cavity.  This is because the magnetic field sensor uses current in motion inside the sensor to sense the dipole redistribution of other currents (the magnetic field) (i.e. dipole electric fields and magnetic fields both decrease by 1/r^3) (see paper 1). 

I would like to propose this as a possible test of the Woodward effect if indeed it is just dealing with time retarded information.  Dual resonant cavities (paper 1) The file is attached below as "magnetic propulsion.pdf".  I think it has potential and appears simple enough to allow understanding of what is happening.  Power in simplicity.

There is a possibly related patent I mentioned earlier but does not deal with resonant cavities though should be possible with radio frequencies and dielectrics: http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&p=1&u=%2Fnetahtml%2FPTO%2Fsearch-bool.html&r=4&f=G&l=50&co1=AND&d=PTXT&s1=electromagnetic.TI.&s2=propulsion.TI.&OS=TTL/electromagnetic+AND+TTL/propulsion&RS=TTL/electromagnetic+AND+TTL/propulsion

Please pardon the horribly long link. 

I also believe DavidWaite is also onto the same thing and you can see his video here:
In his case he flips the solenoid and the static electric field so that they both provide propulsion in the same direction but I suspect in his case you are also dealing with emitted radiation.

If I had the money or connections I would be trying to develop this myself but I am not that fortunate and I don't want to see this possibility pass us by.  I'd at least like to make the connections to see if it will really work.

Offline Cinder

  • Full Member
  • ****
  • Posts: 779
  • Liked: 229
  • Likes Given: 1077
Re: Woodward's effect
« Reply #502 on: 09/26/2015 09:50 pm »

Please pardon the horribly long link. 
https://goo.gl/

Mods can delete this post once that's fixed...

NEC ULTIMA SI PRIOR

Offline dustinthewind

  • Full Member
  • ****
  • Posts: 902
  • U.S. of A.
  • Liked: 313
  • Likes Given: 355
Re: Woodward's effect
« Reply #503 on: 09/29/2015 04:28 pm »
Dustinthewind, maybe you should try to join Woodward's mailing list?


I contacted Woodward myself by e-mail and the impression I got is that he seemed discouraged anything greater than photon propulsion was possible unless gravity was involved.  In my case I would suspect that we might find that it is involved but I would have to probably solve some time dependent field solution of Einsteins field equations which include the electric and magnetic fields.  That would take some practice and research on my part.  One guy I suspect is good at Einsteins field equations relating to electricity and magnetism, and I tried contacting him about it, was WaiteDavidMSPhysics on youtube whose video i posted.  I am not sure he is interested.  I am not sure what Woodward thinks of the EM drives and the apparent forces they get from those yet.  Some of those forces I am sure are convection currents of course but there was one vacuum test by NASA. 
Follow the science? What is science with out the truth.  If there is no truth in it it is not science.  Truth is found by open discussion and rehashing facts not those that moderate it to fit their agenda.  In the end the truth speaks for itself.  Beware the strong delusion and lies mentioned in 2ndThesalonians2:11.  The last stage of Babylon is transhumanism.  Clay mingled with iron (flesh mingled with machine).  MK ultra out of control.  Consider bill gates patent 202060606 (666), that hacks the humans to make their brains crunch C R Y P T O. Are humans hackable animals or are they protected like when Jesus cast out the legion?

Offline Mach1

  • Member
  • Posts: 1
  • Liked: 0
  • Likes Given: 0
Re: Woodward's effect
« Reply #504 on: 09/29/2015 07:26 pm »
Dustinthewind, maybe you should try to join Woodward's mailing list?


I am not sure what Woodward thinks of the EM drives and the apparent forces they get from those yet.  Some of those forces I am sure are convection currents of course but there was one vacuum test by NASA.

Hi there.

This is my first post on this forum. Please bear with me, as I am a layman as far as physics is concerned, but a layman who has followed Woodward's work with interest for years.

With regards to your question in the post above, in his recent (as in a few months ago) interview with the Space Show, Woodward was asked about the EMdrive, and my understanding was that he was fairly dismissive of it, without being impolite.

He seems to be fairly sure that the EMdrive researchers are barking up the wrong tree, and that other than accidental Mach Effects that may be generated in their device, it will ultimately prove to be a futile endeavour.

At least, that was my take from listening to the interview.



Offline flux_capacitor

  • Full Member
  • ****
  • Posts: 708
  • France
  • Liked: 860
  • Likes Given: 1076
Re: Woodward's effect
« Reply #505 on: 09/29/2015 07:45 pm »
Let me summarize this, as I've read what Woodward thinks about the EmDrive: he does not believe that an EmDrive without a dielectric within can work at all; and while he thinks an EmDrive with an internal electrostrictive dielectric could work because of some Mach effect, he denies the reality of any quantum vacuum plasma-based propulsion.

Offline aceshigh

  • Full Member
  • ****
  • Posts: 792
  • Liked: 269
  • Likes Given: 22
Re: Woodward's effect
« Reply #506 on: 09/30/2015 12:45 am »
I was just going to tell DustInTheWind to listen to Woodward's interview on the Space Show, when I saw Mach1 post.


anyway, here are the links

James Woodward interview
http://thespaceshow.com/detail.asp?q=2509

Heidi Fern interview
http://thespaceshow.com/detail.asp?q=2545

Offline Mezzenile

  • Full Member
  • *
  • Posts: 130
  • Liked: 63
  • Likes Given: 24
Re: Woodward's effect
« Reply #507 on: 11/06/2015 04:39 am »
Has James Woodward confirmed that after a sufficient time of operation, the acquired kinetic energy of his standalone device (observed from the initial referential) will overpass the total amount of energy used to operate it ?
« Last Edit: 11/07/2015 07:29 am by Mezzenile »

Offline flux_capacitor

  • Full Member
  • ****
  • Posts: 708
  • France
  • Liked: 860
  • Likes Given: 1076
Re: Woodward's effect
« Reply #508 on: 11/06/2015 09:34 pm »
Has James Woodward confirmed that after a sufficient time of operation, the acquired cinetic energy of his standalone device (observed from the initial referential) will overpass the total amount of energy used to operate it ?

Not at all: Jim Woodward strongly says Mach Effect Thrusters do NOT violate energy and momentum conservation, and can't become overunity free-energy machines. Let me summarize this thought below.

The mistake starts noting that a steady input of power into an ideal Mach effect device should produce a steady thrust.  So the energy required to produce that thrust for some specified length of time is just the power times the length of time it is supplied.  Now, power is dE/dt, so this point of view amounts to the assertion that:

    dE/dt = K F                                                                (1)

where F is the force corresponding to the thrust and K is a constant of proportionality that we can set equal to one with suitable choice of units.  Notice that this relationship can be rearranged to:

    dE = F . dt                                                                 (2)

which, as a mechanical statement, is wrong.  The correct relationship between energy and force in a mechanical system is dE = F . ds

We set aside the fact that Equation (2) is wrong, for though it is mechanically incorrect, and even though proceed with this argument.

Consider the motion of a massive object to which this thruster is attached.  What is the equation of motion?  Newton's second law:  F = ma = m dv/dt.  We can rearrange the equation of motion to:

    dv = (F/m) . dt                                                         (3)

Note that Equation (3) can also be stated as dp = m dv = F . dt, the "impulse" version of the second law.  When F and m are constants, as they are in simple circumstances, Equation (3) can be integrated to:

    v = (F/m) t                                                               (4)

v and t in this equation are to be understood as the final minus initial velocity and final minus initial time.  That is, velocity is a steadily increasing (linear) function of elapsed time when a steady force accelerates an object.

Now some people state that the kinetic energy acquired by the body as it is accelerated is proportional to v^2.  Given the linear dependence of velocity and energy input to the thruster on time, and the fact that the kinetic energy of the body depends on the square of the velocity, thinking so would point out that eventually, no matter what the detailed conditions supposed are, the kinetic energy of the body must exceed the energy put into the thruster to produce the acceleration.  Therefore, you would argue that Mach effect thrusters violate energy conservation.  Then you would believe that you should reject Mach effects because of their alleged violation of a well-known conservation principle.

But this argument is wrong, and Woodward and others have tried to explain why. Others have tumbled to this flawed argument and tried to spread it as well.  Please note that this argument doesn't simply apply to Mach effect systems (of which you may be suspicious).

Consider a rocket motor attached to a rotor arm. (This example can be done with linear motion, but then you have to invoke a lot improbable ideal apparatus) Fuel is delivered to the motor through tubes in the rotor arm at a steady rate -- satisfying the condition of steadily delivered energy to the motor. The rotor is in a vacuum and the bearings are frictionless.  The motor fires and it is accelerated at a constant rate. (We're ignoring the vector properties of acceleration and velocity here, and the centripetal acceleration that keeps the motor from flying off in a straight line. The usual vector properties would obtain if we were doing the linear example.) The velocity of the motor increases as a linear function of time. The kinetic energy of the motor is (1/2) m v^2, so it increases as the square of the time, just as in the Mach thruster case.  Guess what? Eventually the kinetic energy of the motor exceeds the input energy in the fuel. Another violation of energy conservation!

Now, since that specious argument leads to energy conservation violations for both Mach systems and easily envisaged normal systems, it should be obvious that the problem is NOT that energy conservation is being violated.

The correct way to do the calculation is to consider very small increments of time -- increments so small that the kinetic energy acquired by the accelerating body in the interval remains a small fraction of the input energy.  One then sums the input and kinetic energies for the intervals over the duration of the application of the thrust. Since the kinetic energy never exceeds the input energy in any interval, the summed kinetic energy increments cannot exceed the input energy increment sum. No conservation violation. Reasonable physics.
« Last Edit: 11/06/2015 09:36 pm by flux_capacitor »

Offline Paul451

  • Senior Member
  • *****
  • Posts: 3553
  • Australia
  • Liked: 2518
  • Likes Given: 2180
Re: Woodward's effect
« Reply #509 on: 11/07/2015 01:46 am »
The correct way to do the calculation is to consider very small increments of time -- increments so small that the kinetic energy acquired by the accelerating body in the interval remains a small fraction of the input energy.  One then sums the input and kinetic energies for the intervals over the duration of the application of the thrust. Since the kinetic energy never exceeds the input energy in any interval, the summed kinetic energy increments cannot exceed the input energy increment sum. No conservation violation. Reasonable physics.

Uh, no. If you use a consistent frame of reference, then the rate of change of velocity and rate of change of energy are instantaneous measures, not cumulative. At any velocity above 1/alpha - where alpha is the devices efficiency in newtons of thrust per watt of input power - increase in output energy exceeds rate of energy input. Your objection doesn't hold.

Similarly...

Consider a rocket motor attached to a rotor arm. [...] Fuel is delivered to the motor through tubes in the rotor arm at a steady rate [...]
Guess what? Eventually the kinetic energy of the motor exceeds the input energy in the fuel. Another violation of energy conservation!

No, the angular momentum of the fuel being fed through the arm needs to increase from zero at the hub, to r*m*v at the rocket - this creates a decelerating force on the arm that is also proportional to tangential velocity. The net energy output therefore never exceeds the energy content of the fuel.

This applies to all conventional systems, in different ways, there's always a second system you need to account for. A jet/prop interacts with the medium it's pushing against. A magnetic/electrical system interacting with a fixed field. Etc etc.

The only exception is a perfect photon drive. However the velocity above which a photon emitter gains energy faster than it consumes it is exactly 'c'.

With a reactionless thruster, such as an MET or EMDrive, the overunity velocity occurs at much more mundane speeds. At the oft quoted alpha of 4 newtons per kilowatt, the overunity velocity is just 250 m/s.

Now, since that specious argument leads to energy conservation violations for both Mach systems and easily envisaged normal systems, it should be obvious that the problem is NOT that energy conservation is being violated.

And since your argument was wrong... it should be obvious that... ?

Offline 93143

  • Senior Member
  • *****
  • Posts: 3054
  • Liked: 312
  • Likes Given: 1
Re: Woodward's effect
« Reply #510 on: 11/07/2015 04:44 am »
I have already shown elsewhere that a rocket of any type cannot go over unity, whether or not it is mounted on a flywheel.  I have also derived a condition that must apply to the Mach effect thruster if it works the way the equations seem to say it does (ie: no dependence of thrust on velocity) if global conservation of energy is to be respected.

http://talk-polywell.org/bb/viewtopic.php?f=10&t=2215&p=103524#p103524
http://talk-polywell.org/bb/viewtopic.php?f=10&t=2215&p=103729#p103729
http://talk-polywell.org/bb/viewtopic.php?f=10&t=2215&p=105085#p105085

Essentially, the effective mean velocity of the far-off active mass being interacted with must be invariant with respect to the thruster in order for energy to be conserved in all reference frames.  The most plausible-seeming solution that presents itself is that the interaction happens in such a way (relativistic Doppler effect, perhaps?  Something more esoteric?  I need to catch up on the literature) that the effective mean velocity of the far-off active mass is always equal to that of the thruster.

But we can dispense with the nonsense about critics doing the math wrong.  Some do, but the upshot is that if the M-E thruster works as advertised, you can make something that looks exactly like a perpetual motion machine if you ignore the interaction with the rest of the universe.

That interaction with the rest of the causally-connected universe is where the "extra" energy comes from.  In fact it is the entire reason anything happens at all.  The work done by an M-E thruster is largely unrelated to the local energy input, in the same sense in which the work done by the wind on a sailboat is largely unrelated to the energy expended by the crew moving the sails around.  As far as I know there is no theoretical upper limit on the thrust efficiency of a Mach-effect device.
« Last Edit: 11/07/2015 07:33 am by 93143 »

Offline Mezzenile

  • Full Member
  • *
  • Posts: 130
  • Liked: 63
  • Likes Given: 24
Re: Woodward's effect
« Reply #511 on: 11/07/2015 06:12 am »
Has James Woodward confirmed that after a sufficient time of operation, the acquired cinetic energy of his standalone device (observed from the initial referential) will overpass the total amount of energy used to operate it ?
Not at all: Jim Woodward strongly says Mach Effect Thrusters do NOT violate energy and momentum conservation, and can't become overunity free-energy machines. Let me summarize this thought below ...
Sorry but your main example of a conventional rotary thruster device does not resist to the analysis of Paul451. As Paul said, you have missed the point that energy is required to bring the propelant at the speed of the thruster nozzle before it can produce its thrust !

I raised this question on kinetic energy because I have not managed to find in Woodward papers any reference to this subject. For me this is a bit surprising ...

I know that the General Relativity of Einstein does not impose a conservation of energy at the level of the universe and has even great difficulty to define a local conservation of energy when gravitational tidal forces are acting. So I don't think that the Mach hypothesis is challenged by this phenomena. Simply  I was expecting Woodward to comment this rather amazing consequence of his Mach thruster discovery  !  :)
   

Offline Mezzenile

  • Full Member
  • *
  • Posts: 130
  • Liked: 63
  • Likes Given: 24
Re: Woodward's effect
« Reply #512 on: 11/07/2015 07:58 am »
I have already shown elsewhere that a rocket of any type cannot go over unity, whether or not it is mounted on a flywheel.  I have also derived a condition that must apply to the Mach effect thruster if it works the way the equations seem to say it does (ie: no dependence of thrust on velocity) if global conservation of energy is to be respected.

http://talk-polywell.org/bb/viewtopic.php?f=10&t=2215&p=103524#p103524
http://talk-polywell.org/bb/viewtopic.php?f=10&t=2215&p=103729#p103729
http://talk-polywell.org/bb/viewtopic.php?f=10&t=2215&p=105085#p105085

...
...

That interaction with the rest of the causally-connected universe is where the "extra" energy comes from.  In fact it is the entire reason anything happens at all.  The work done by an M-E thruster is largely unrelated to the local energy input, in the same sense in which the work done by the wind on a sailboat is largely unrelated to the energy expended by the crew moving the sails around.  As far as I know there is no theoretical upper limit on the thrust efficiency of a Mach-effect device.
Thank you for the links.  :)

Concerning energy and General Relativity there is a theorem which says that the total energy of an isolated system cannot be negative (theorem of the 80's). So this seems to bounds the amount of energy that can be radiated away in the form of gravitational radiation from any closed system. May be this can at least prevent our Mach thruster to pump an infinite energy from the remaining whole universe  by putting it in a state of more and more local negative energy  :).  But whatever, this let great margins to do very interesting things with the device of Woodward  !  ;)

Offline 93143

  • Senior Member
  • *****
  • Posts: 3054
  • Liked: 312
  • Likes Given: 1
Re: Woodward's effect
« Reply #513 on: 11/07/2015 08:11 am »
Just to be extra clear (since some people have trouble with this sort of thing), this all depends on the M-E thruster actually working as claimed, which as far as I know does not yet appear to be beyond reasonable doubt.

If it does work, yes, it'll be great.

Also, please note that since the thruster is supposed to be interacting with the distant universe via gravinertial transactional radiation, it is not an isolated system...
« Last Edit: 11/07/2015 08:13 am by 93143 »

Offline Mezzenile

  • Full Member
  • *
  • Posts: 130
  • Liked: 63
  • Likes Given: 24
Re: Woodward's effect
« Reply #514 on: 11/07/2015 08:27 am »
Just to be extra clear (since some people have trouble with this sort of thing), this all depends on the M-E thruster actually working as claimed, which as far as I know does not yet appear to be beyond reasonable doubt.

If it does work, yes, it'll be great.

Also, please note that since the thruster is supposed to be interacting with the distant universe via gravinertial transactional radiation, it is not an isolated system...

You right, it should be made clear what is the definition of an isolated system in the theorem I refer to,  as in term of Machian gravitation, at soon as an object is accelerated it seems to interact with the whole causally connected universe.

I imagine also that the energy transfer from the Universe to a Mach thruster must have an impact on the entropy side and so have thermodynamic consequences. Here also I imagine that Woodward has something to talk about ...
« Last Edit: 11/09/2015 02:53 pm by Mezzenile »

Offline Paul451

  • Senior Member
  • *****
  • Posts: 3553
  • Australia
  • Liked: 2518
  • Likes Given: 2180
Re: Woodward's effect
« Reply #515 on: 11/07/2015 08:53 am »
Also, please note that since the thruster is supposed to be interacting with the distant universe via gravinertial transactional radiation, it is not an isolated system...

Too much of a handwave explanation. Similar to Shawyer objecting to the people calling his device "reactionless" because he put a second arrow on his diagram labelled "reaction".

Offline 93143

  • Senior Member
  • *****
  • Posts: 3054
  • Liked: 312
  • Likes Given: 1
Re: Woodward's effect
« Reply #516 on: 11/07/2015 08:01 pm »
The whole point of this device is that it works because Mach's Principle is true.

And if Mach's Principle is true (as suggested by the correspondence between the theoretical requirements and recent cosmological measurements), then it's not just M-E thrusters that work via interaction with the distant universe - it's inertia itself.

The mechanism has been described as quadrupole gravity radiation generated by the system consisting of the object in question and the rest of the causally-connected universe.  The transactional bit might have been a bit of a handwave (though it made sense and had a theoretical precedent), but apparently Hoyle-Narlikar gravity (slightly modified to harmonize it with modern cosmology) explicitly describes this mechanism.  So there is a theoretical basis.

This is not at all like drawing an extra arrow on a free body diagram to obscure the fact that your idea violates Newton's Third Law.

Offline Katana

  • Full Member
  • ***
  • Posts: 378
  • Liked: 49
  • Likes Given: 20
Re: Woodward's effect
« Reply #517 on: 11/08/2015 02:42 pm »
Breaking basic physical laws should be a topic of theoretical physics instead of aerospace engineering.

Have anybody asked relevent questions to Stephen Hawking?

Offline birchoff

  • Full Member
  • **
  • Posts: 273
  • United States
  • Liked: 125
  • Likes Given: 95
Re: Woodward's effect
« Reply #518 on: 11/08/2015 08:35 pm »
Breaking basic physical laws should be a topic of theoretical physics instead of aerospace engineering.

Have anybody asked relevent questions to Stephen Hawking?

Have you done your homework and reviewed all of the published papers from Woodward, and more recently heidi fearn?

[edited]

I am not trying to be confrontational. But unlike the EmDrive Woodwards Mach Effect is being developed by people who do theoretical physics, from my perspective. assuming you havent been aware of the latest published papers I would refer you to the following link. You can follow the references to get an understanding of the theory behind the effect they are observing in experiments.

http://www.researchgate.net/publication/280134421_New_Theoretical_Results_for_the_Mach_Effect_Thruster
« Last Edit: 11/08/2015 08:44 pm by birchoff »

Offline birchoff

  • Full Member
  • **
  • Posts: 273
  • United States
  • Liked: 125
  • Likes Given: 95
Re: Woodward's effect
« Reply #519 on: 11/08/2015 11:24 pm »
Theory of a Mach Ef ect Thruster 2 is now out published in Journal of Modern Physics just like Part 1

Theory of a Mach Effect Thruster I

Theory of a Mach Effect Thruster II

from my first read it looks...

1. Woodward and Fearn now have a derivation of Woodward's force equation which he derived from GR equations using HN Theory.
a. Also derived the mass fluctuation equations from HN-Theory and transformed it to be more engineer friendly

2. Also seem to have a better justification for why the scaling will be w^3/w^4 instead of w^6. Compared to the gut feeling expectation discussed in Part 1

3. Also have a much clearer explaination for not only how the mass fluctuations manifest but how it is rectifed into a consistent force. Along with the requirements for making it happen.(Definately get the impression that they now have not only the theory down but also the engineering paramters).

4. Also included in their conclusion is that there are two successful replications. One is from Nembo Baldrin in Austria and the other is from someone in Canada. From what I can tell in the paper both replicators used MET's built by James Woodward, Headi Fearn, and Keith Watsner (JFW)

[edit]
I made a mistake I originally thought part 1 was the paper headi presented at the AIAA conference earlier this year. Where as Part 1 is beginning of what she promised at the end of her AIAA paper which is a complete derivation of Mach effects from HN Theory.
« Last Edit: 11/08/2015 11:36 pm by birchoff »

Tags:
 

Advertisement NovaTech
Advertisement Northrop Grumman
Advertisement
Advertisement Margaritaville Beach Resort South Padre Island
Advertisement Brady Kenniston
Advertisement NextSpaceflight
Advertisement Nathan Barker Photography
0