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)
QuoteThe 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.
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:
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.
Quote from: ppnl on 07/21/2018 03:57 amImagine 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.
Quote from: Bob Woods on 07/21/2018 04:42 amQuote from: ppnl on 07/21/2018 03:57 amImagine 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.
"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.
Quote from: soms42 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.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.
Quote from: ppnl on 07/22/2018 02:17 amQuote from: soms42 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.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 "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.
QuoteThe "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?
Quote from: soms42 on 07/23/2018 07:16 pmQuote from: ppnl on 07/22/2018 02:17 amQuote from: soms42 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.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.
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.
QuoteThe 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.
With the frame issue, its almost like we're getting into an area where even ordinary rockets can't work.
No, the MET still has several problems here that lead to local power generation.