A FINAL WORD on the thermodynamics of 'Mach Thrusters'.

Discuss life, the universe, and everything with other members of this site. Get to know your fellow polywell enthusiasts.

Moderators: tonybarry, MSimon

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm

Post by chrismb »

I don't understand what you mean "F*(vr-vu) doesn't work". That is by definition the total rate of change of kinetic energy. Maybe you are confused because Fu is negative: F = Fr = -Fu.[/quote]
Exactly: F*vu cannot be negative, if it is defining a KE increase.

kcdodd
Posts: 722
Joined: Tue Jun 03, 2008 3:36 am
Location: Austin, TX

Post by kcdodd »

Um, by definition P = Fr*vr + Fu*vu. If you disagree with this, then I don't know how else to explain it. Go sit down and derive total rate of change of kinetic energy and show me the correct equation for total power. If I substitute in Fu = -Fr, then I get P = Fr*vr - Fr*vu = Fr*(vr - vu). Where is the problem.
Carter

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm

Post by chrismb »

kcdodd wrote:Um, by definition P = Fr*vr + Fu*vu. If you disagree with this, then I don't know how else to explain it. Go sit down and derive total rate of change of kinetic energy. If I substitute in Fu = -Fr, then I get P = Fr*vr - Fr*vu = Fr*(vr - vu). Where is the problem.
What does kcd think those equation at the top of the thread are!!??

Derived by both change of KE and by work performed!!!

This 'moving' uv term is the thing that screws it in regards the guy flying the rocket. As far as he is concerned, when he accelerates it is the Mu that accelerates!

As far as an observer who was stationary wrt Mu before the thrust started, he would see the universe get thrusted one way and the thruster the other. He would add up those two KEs. Clearly - very clearly - if he sees an object accelerate from 1000m/s to 1010m/s then it would be a lot more KE than if another observer saw the same acceleration as if it were from 0m/s to 10m/s.

Just think about it for a moment - is the observed change of KE really the same for a given velocity increment observed in different frames?

This would seem to be getting to the bottom of the fallacy that is described as being elsewhere.

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm

Post by chrismb »

kcdodd wrote:Um, by definition P = Fr*vr + Fu*vu.
Well, it's not entirely wrong but easy to get confused on the signs. Remember
Fr = -Fu,
and vr = -vu.(Mu/Mr)

So these add up to a double negative, so the two terms add.

Now;
F*vr > F*v [because v is taken as the velocity of the rocket instantaneously after acceleration has begun, whilst vr is this 'real' velocity], so
F*vr-F*vu >> F*v [because that vu is negative wrt vr, so the LHS is bigger than it was in F*vr > F*v].

kcdodd
Posts: 722
Joined: Tue Jun 03, 2008 3:36 am
Location: Austin, TX

Post by kcdodd »

chrismb wrote:
kcdodd wrote:Um, by definition P = Fr*vr + Fu*vu.
Well, it's not entirely wrong but easy to get confused on the signs. Remember
Fr = -Fu,
and vr = -vu.(Mu/Mr)

So these add up to a double negative, so the two terms add.

Now;
F*vr > F*v [because v is taken as the velocity of the rocket instantaneously after acceleration has begun, whilst vr is this 'real' velocity], so
F*vr-F*vu >> F*v [because that vu is negative wrt vr, so the LHS is bigger than it was in F*vr > F*v].
No, F*vr - F*vu = F*v. Which is what I wrote initially: P0 = Pk. v is defined as the relative velocity, v = vr - vu. It's direct substitution. Use the algebra. That's why it's there. You're not using the defintions. v is bigger than vr by a teeny tiny amount, not smaller than vr.
Carter

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm

Post by chrismb »

OK. Looks like these expressions are intending to be saying the same thing, just missing the common language describing them. Apologies due to kcd.

Looks like a chrismb mis-comprehension issue with the terminology 'what the pilot reads from the engine'. The pilot's frame is accelerating and already moving wrt Mu, as a result his engines get more 'leverage', as it were, from the increasing differential velocities (vr) as he accelerates, so any observed vu changes come with a KE 'offset'. This then balances out, when looked at from any non-accelerating perspective, so long as the mu KE isn't forgotten about.

This is akin to the propellant in a conventional rocket gaining an apparent 'KE advantage' as the rocket fuel itself gets faster, thus the deceleration once ejected increases relatively to some independent inertial frame.

This is probably at the heart of the 'apparent paradox' folks have discussed, but actually there isn't any. Like the mis-aimed discussion in the last several posts. It is simply that sometimes one object looks to get the KE 'power' and the other gets none (@ t=0), or vice versa, dependent of the frame of reference. This might certainly lead to some confusions. Change the frame of reference (e.g. relative to an acceleration, or swap a 'KE' between two frames) and all the maths turns to doo doo.

kcdodd
Posts: 722
Joined: Tue Jun 03, 2008 3:36 am
Location: Austin, TX

Post by kcdodd »

For me, the more disconcerting aspect is that the power consumption of the engine comes out dependent on the relative motion v = vr - vu. So, just like a hybrid car can take it's own kinetic energy and turn it back into stored energy. If you hit the thruster in "reverse", then F -> -F, and the power goes negative. So, any time the thruster pushes in opposite direction as v (the relative motion wrt to the universe COM velocity). I don't necessarily think this is a violation of 2nd law of thermodynamics (or the first), but it just doesn't seem likely at all.

So, I have serious doubts that such a thruster could be physically real. The closest would be some kind of gravitational wave drive, which would be no more efficient than a light thruster, or shining a flashlight out the back of the spaceship.
Carter

CaptainBeowulf
Posts: 498
Joined: Sat Nov 07, 2009 12:35 am

Post by CaptainBeowulf »

Chris, my link on page 2 may be awful from a formatting point of view, but it goes directly to the chapter in question, whereas the shortened link you tried seems to just go to google's overview page for the book, instead of to the precise page. I blame the internet.

(Incidentally, the long link does seem to wrap on the two computers I've looked at it with today, so it must also depend on the browser edition one is using. It does make the screen far too wide in my smartphone, though.)

MSimon
Posts: 14332
Joined: Mon Jul 16, 2007 7:37 pm
Location: Rockford, Illinois
Contact:

Post by MSimon »

How about this:

Barbour
Engineering is the art of making what you want from what you can get at a profit.

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm

Post by chrismb »

kcdodd wrote:... the power consumption of the engine comes out dependent on the relative motion v = vr - vu. So, just like a hybrid car can take it's own kinetic energy and turn it back into stored energy. If you hit the thruster in "reverse", then F -> -F, and the power goes negative. So, any time the thruster pushes in opposite direction as v (the relative motion wrt to the universe COM velocity). I don't necessarily think this is a violation of 2nd law of thermodynamics....
Yup. It's what the maths at thread top says.

So, with this in mind, the question of whether this alludes to a means to testing whether a reaction force has been accomplished with ROTU can be looked at:

If an oscillator is driven by a conventional sinus force derived in the frame of the centre of harmonic motion, then one sees an applied (restoring) force at the ends of the harmonic travel, Ө=90° & 270° and no force applied to the oscillating mass as it passes through Ө=180° & 360°, with the profile of the force (and applied power) being symmetrical about Ө=180° to achieve a similar symmetrical profile for the motion.

Now imagine a mass is oscillating back-and-forth in a moving vehicle, and, somehow(!?!?), the force driving that oscillation isn't in the frame of the moving vehicle but is from the road it is driving along. For the mass to be driven forward of Ө=0°, relative to the motion of the vehicle it has bigger differential KE changes going on than if it were being slowed and falling behind Ө=0°, viz. Ө=270° (taking that to be its slowest 'real' speed) the instantaneous power demand there would be lower than at Ө=90°.

So if an oscillator driver is from any 'leveraged' power from the fabric of the universe then for a symmetrical application of power one would get an asymmetrical displacement versus time, or similarly an observation of symmetrical motion would come from an asymmetric application of power.

The differential asymmetry would be a function of the relative speed wrt this 'ether' medium. If the relative speed is zero, then it is in the same frame and it would be all symmetric, because the oscillation is then in the same frame.

As the differential velocity increases, one will see the asymmetry increase.

If one performs the same experiment but facing the opposite direction, the asymmetry in the harmonic motion should reverse.

There would be no need to reconfigure and reverse the experiment to examine this - instead, simply, just wait 12 hours!

Therefore, one must see an asymmetry in the power demand to achieve a symmetrical SHM for a propellantless thruster at some point during the day, and that the degree of asymmetry would be observed to vary diurnally.

If there is no diurnal variation in power consumption cyclic symmetry when powering an oscillating mass then there is no reaction to some feature of the wider universe.

An observation of diurnal variation in harmonic power consumption symmetry, to achieve a symmetrical SHM, would be consistent with some extant reaction force into another remote frame.

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm

Post by chrismb »

It may also be noted that the last post is both for a general case, and also in the eventuality that someone says that a force generated either way would cause a power demand from a propellantless thrust.

Of course, the much more trivial observation is that during conventional SHM with forces derived in the frame of the mass at Ө=0°, then there is:
power dissipation, from Ө=0° to 90° (mass slows)
power demand, from from Ө=90° to 180° (mass accelerates)
power dissipation, from Ө=180° to 270° (mass slows)
power demand, from from Ө=270° to 360° (mass accelerates)

Whereas for a mass accelerating by forces derived in a frame >peak SHM velocity*, then;
power dissipation, from Ө=0° to 180° (mass slows)
power demand, from from Ө=180° to +360° (mass accelerates)

So an oscillator driven by forces in the inertial frame of its Ө=0°, the power demands will occur at twice the oscillation frequency, whereas an oscillator driven by external 'far-off' forces in a frame faster than its velocity* at Ө=0° would be at the oscillation frequency.

*[careful not to discount the relative motion of the Earth in that condition - hence take the asymmetric power demand as the more deterministic approach]

In other words, a system driven by power cycles** of frequency 2F that oscillates in SHM at F is bound in the same frame as the power, whereas if the SHM oscillator runs at 2F and oscillates at 2F then this is consistent with a force connection to some fabric of the universe.

**(remember that there are two power cycles in one AC cycle. Also to note - if there is an elastic/gravity/energy storage system then that may be additional power cycles each time the energy reserve operates - viz. force a mass forward against a spring, then release the force so the spring returns - this is two power cycles, within the definition of '2F' above)

kcdodd
Posts: 722
Joined: Tue Jun 03, 2008 3:36 am
Location: Austin, TX

Post by kcdodd »

This is sort of back to the result of either theory, with a preferred reference frame, which has been debunked. So it's hard to imagine anything would work this way.
Carter

Post Reply