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 »

CaptainBeowulf wrote:Does the universe have a rotational angular momentum? No (at least assuming Mach's Principle is true).
The rotation thing introduces a whole new dimension. The analogy of a car on the surface of the earth was not intended to suggest rotation was the end result. However, the analogy is significant indeed.

The above analysis is a purely 1D analysis - all momentum change in the same line. One can instead think of a long geared track in space of considerable mass, along which a geared cart accelerates. Same things apply, no rotation.

But if the CoG of the ROTU is, indeed, not in line with the direction of acceleration of a propellantless thruster then that thruster would gain rotational momentum wrt ROTU if it were to accelerate. Conservation of momentum would dictate that the ROTU would then have to acquire rotational energy.

So;
if momentum is transferred from a prospective propellantless thruster to the ROTU,
AND
if the acceleration of the thruster is not inline with the CoG of the universe
AND
if the ROTU cannot rotate
THEN
the thruster cannot accelerate, because the only solution that satisfies the above statements is zero rotational momentum change in both ROTU and thruster.

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

Post by chrismb »

CaptainBeowulf wrote:Does the universe have a momentum in a certain direction? No. However, mass in the universe does have a momentum - it's expanding outwards because some expansionary force is stronger than gravity.
This makes little sense. Either it has a momentum in a direction (which may be a zero momentum) but if it is 'undefined' and momentum is transferred to it somehow by another object then its momentum would change, so cannot be undefined.

Incidentally, just because an ensemble of objects is expanding away from each other does not mean they have a net momentum. Think of an explosion inside a strong box that can withstand the explosion. The box and its contents does not gain any momentum as a result of explosions going off inside it.

If (and it has been bandied around so much that it seems jaw-droopingly staggering that GIT would now suggest some alternative thing) momentum is transferred from a thruster to ROTU by means yet to be understood, then the remaining constituents of the universe, the ROTU except the thruster, must then change their momentum. Clearly, like the passenger trying to push the bus from inside, there is no actual change of momentum for the whole universe (ROTU+thruster), but energy will still be expended, or dissipated, nonetheless so as to change the relative velocity between the ROTU and the thruster.

GIThruster
Posts: 4686
Joined: Tue May 25, 2010 8:17 pm

Post by GIThruster »

So;
if momentum is transferred from a prospective propellantless thruster to the ROTU,
AND
if the acceleration of the thruster is not inline with the CoG of the universe
AND
if the ROTU cannot rotate
THEN
the thruster cannot accelerate, because the only solution that satisfies the above statements is zero rotational momentum change in both ROTU and thruster
No chris. And AGAIN we have an example of you posting what you think makes sense as a critique when all you're doiong is acting confused because you have not done your homework. Read the paper for cryin' out loud! It is not worth anyone's time to correct you in these assinine objections and statements when it is your laziness and stupidity that is their cause.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

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

Post by chrismb »

GIT has a capacity for reasoning that is unchallengeable.

OK, so he has no reasoning capability, but nonetheless it is unchallengeable!! :lol:

Skipjack
Posts: 6110
Joined: Sun Sep 28, 2008 2:29 pm

Post by Skipjack »

Hmm, not sure how that affects the thinking process, but I have been wondering whether it is better to imagine a swimmer diving under water, pushing himself forward with his hands. In a sense the entire pool is his reactionmass (though in this case the nearby water molecules are receive the most acceleration). Another comparison could be a subarine diving.
So how does that compare to Woodwards ideas?

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

Post by chrismb »

Skipjack wrote:Hmm, not sure how that affects the thinking process, but I have been wondering whether it is better to imagine a swimmer diving under water, pushing himself forward with his hands. In a sense the entire pool is his reactionmass (though in this case the nearby water molecules are receive the most acceleration). Another comparison could be a subarine diving.
So how does that compare to Woodwards ideas?
This thread is [despite GIT's outrageous claims] a simple analysis of a starting posit, which is that a force can be applied between an object and the rest of the universe. Irrespective of the veracity/physicality of that, with that as a starting point the rest follows. It makes no claim to compare with any other theory.

Beyond that is beyond any conclusions that such a posit would lead to. In the case of a swimming pool, it's an interesting alternative that does still fit in with the analysis, but is indeed probably closer to what is envisioned for such proposed thrusters because the swimmer's momentum is transferred to the water's net momentum, so the net momentum of swimmer and water does not need to be considered to change.

That analogy is fine, and leads, again, to the inevitable numerical conclusion above. Imagine that rather than a swimmer it is a jet-ski (easier to think of). Here, as the jet ski gets faster and faster it has to chuck the water out faster and faster to sustain the same thrust, or in other words, the thrust from a given flow rate of water scooped up drops off very fast as the thing picks up speed.

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

Post by kcdodd »

I prefer to keep this type of thing in terms of thought experiments. If you can't connect this back to reality in some way, then it's pointless. And that is what I was getting at with my comment about a bucket with a hole in the bottom.

Imagine I'm just out floating in space, in a space suit, and there is a rocket near by with an engine that ejects nothing, but produces thrust accelerating the rocket. So, what are you going to see when it starts up? Assume I can perform measurements of distance and time somehow (ie I can measure x, v, a, t, etc).

Knowing nothing else, I would expect that the rocket accelerates at a = F/m. And the kinetic energy of the rocket increases at a rate of Pk = F*v. This is because I assume I am in an inertial reference frame: I feel no forces on myself, and don't appear to accelerate relative to the rest of the universe.

Now I get on skype and start chatting with the guy at the helm, that reads off the power consumption of the engine to me, let's say P0. That is the big question, isn't it? Is P0 = Pk, P0 > Pk, or P0 < Pk?

From my earlier post, I was thinking the first one is out, because in reality, I, and the rest of the universe, are supposedly being pushed on by this engine even though we can't tell it. So whatever the pilot reads off cannot be identically equal to what I see.

But, to say more than this requires more assumptions and more math. I think it's safe to assume that the universe has a small acceleration is opposite to that of the rocket. So the velocity I measure is bigger than the "real" velocity. Now what. If I assume the whole universe accelerates together? Well, this can't be physical because that would be instantaneous forces, but anyway, I don't know what else to do. Mu, universe mass, Mr, rocket mass, then Mu*vu + Mr*vr = 0, where vu is "real" universe average velocity, and vr is real rocket velocity. So the velocity I see is:

v = vr*(1 + Mr/Mu)

Which means the power I see is

Pk = F*v = F*vr*(1 + Mr/Mu)

And the total real power consumed by the rocket (what the pilot reads from the engine) is

P0 = F*(vr - vu) = F*vr*(1 + Mr/Mu)

So indeed Pk = P0. No problem there. That's because even though I see a higher kinetic energy gain of the rocket than in reality, I also don't see the kinetic energy gain of the rest of the universe. So they balance out.
Carter

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

Post by MSimon »

chrismb wrote:GIT has a capacity for reasoning that is unchallengeable.

OK, so he has no reasoning capability, but nonetheless it is unchallengeable!! :lol:
The ability to reason outside what has been taught is zero for most folks past age 25. You can see that on any number of topics around here. If you reason outside a person's ability they become very uncomfortable to the point of hostility.

I get alternately accused of being a non-conformist or a mere follower by the same people on the same topic. I am amused.

There is a difference between understanding and investment. Understanding is subject to change. Investment is only subject to loss. And humans are very loss adverse.
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:Which means the power I see is

Pk = F*v = F*vr*(1 + Mr/Mu)

And the total real power consumed by the rocket (what the pilot reads from the engine) is

P0 = F*(vr - vu) = F*vr*(1 + Mr/Mu)

So indeed Pk = P0. No problem there.
It is possible kcdodd will need to reconsider this maths...

It is saying that v=vr-vu. Well, if the real vu is zero, in the starting frame (and the calculation is based on the instantaneous power at some 't=0' when the rocket lights, and is immediately no longer applicable*), then does v (the observed rocket velocity) always equal the real rocket velocity, and at all times later too?

This is why the partial differential form above is the rigorous approach, because it is only doing the analysis for that delta t whilst vu 'is still zero'. As soon as it is non-zero, the numbers all change.

The 'real' rocket power, in this notation, is simply F*vr. If one is moving in 'roughly' the same frame as the rocket then F*v is low, whilst if that frame is very fast in the 'real' frame then F*vu will be high.

F*(vr-vu) doesn't work in 'linear mechanics' because it is subtracting from the end result the change of KE that the universe is experiencing in the opposite direction. That's "having one's cake and eating it", because the total KE power change 'in the real frame' would be the two KE's added together.

One needs to be very careful in using the power = change of momentum equation. It has to be dealt with for each individual particle, not net momentum change (else it would mean no KE change is ever possible within a discrete system!!!). Imagine the explosion in a sealed box - plenty of KE inside the box, but no change of momentum overall. One would have to write something like ∑√(Fi.vi)² for i=1 to n particles with different velocities (note, velocities are vectors, so particles can be only grouped into an 'n-th' ensemble particle if they have the same magnitude and direction).

Skipjack
Posts: 6110
Joined: Sun Sep 28, 2008 2:29 pm

Post by Skipjack »

chrismb wrote: That analogy is fine, and leads, again, to the inevitable numerical conclusion above. Imagine that rather than a swimmer it is a jet-ski (easier to think of). Here, as the jet ski gets faster and faster it has to chuck the water out faster and faster to sustain the same thrust, or in other words, the thrust from a given flow rate of water scooped up drops off very fast as the thing picks up speed.
I prefer the analogy of a submarine and its propeller. Same problem though. The faster you go, the more water you have to move.
Now my question is: Does Woodwards theory really claim constant acceleration independent of the speed?

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

Post by chrismb »

Skipjack wrote:I prefer the analogy of a submarine and its propeller.
Better still, indeed!

So, now contemplate if some race of beings lived in what appears to be a sealed capsule bobbing around within a medium, and one of them created a gadget that he claimed could thrust against the surrounding medium.

Discount using expelled propellant, because it is the hypothesis that a means exists to do so without propellant.

So to generate thrust against the surrounding medium this inventor would have had to invent something like a propeller, that mechanically pushes against the external media.

OK, so this is now the interesting bit, and is a piece of real science to derive a suitable hypothesis for propellantless thrusters: what tests could the other beings do to prove that he was causing motion of their capsule by the use of a mechanical-type propeller (and not by any other transient means such as rocking the capsule back and forth but not actually producing additional momentum)?

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

Post by MSimon »

Also this. Think of walking on a small very light boat in the water.

You can add momentum to the boat by walking fore to aft (also the other way but it doesn't sound as smooth in English ) but as soon as you stop moving so does the boat. You have changed the center of mass. But there is no net velocity once you stop walking. And of course to change the center of mass chemical energy was expended.
Engineering is the art of making what you want from what you can get at a profit.

Skipjack
Posts: 6110
Joined: Sun Sep 28, 2008 2:29 pm

Post by Skipjack »

I would have regarded the entire universe as an ocean and the ME thruster as the propeller. It seems to make sense if you look at it that way. Of course this would imply that the faster you go, the more energy, you would need.

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

Post by chrismb »

Skipjack wrote: Of course this would imply that the faster you go, the more energy, you would need.
Absolutely. This is what the maths clearly says. And, of course, if one were barrelling along under the high seas at speed, then forcibly slowing oneself down against the sea would dissipate power rather than consume it.

Power dissipation would be through the same means by which 'traction' would be gained - viz. if one were to be shooting along in one's sub and the propellers were clutched on to some generators, then the drag as the generators slow the propeller screw by taking power off them would be the thing that slows the sub (notwithstanding the usual flow resistances - but this is just an analogy to a proposed 'friction-free' system where objects ordinarily move without any resistance in the universal 'ether'/'gravity fields'/whatever).

It's all there, in the maths above .... plain as day.
Last edited by chrismb on Sat Feb 09, 2013 10:16 pm, edited 1 time in total.

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

Post by kcdodd »

chrismb wrote:
kcdodd wrote:Which means the power I see is

Pk = F*v = F*vr*(1 + Mr/Mu)

And the total real power consumed by the rocket (what the pilot reads from the engine) is

P0 = F*(vr - vu) = F*vr*(1 + Mr/Mu)

So indeed Pk = P0. No problem there.
It is possible kcdodd will need to reconsider this maths...

It is saying that v=vr-vu. Well, if the real vu is zero, in the starting frame (and the calculation is based on the instantaneous power at some 't=0' when the rocket lights, and is immediately no longer applicable*), then does v (the observed rocket velocity) always equal the real rocket velocity, and at all times later too?

This is why the partial differential form above is the rigorous approach, because it is only doing the analysis for that delta t whilst vu 'is still zero'. As soon as it is non-zero, the numbers all change.

The 'real' rocket power, in this notation, is simply F*vr. If one is moving in 'roughly' the same frame as the rocket then F*v is low, whilst if that frame is very fast in the 'real' frame then F*vu will be high.

F*(vr-vu) doesn't work in 'linear mechanics' because it is subtracting from the end result the change of KE that the universe is experiencing in the opposite direction. That's "having one's cake and eating it", because the total KE power change 'in the real frame' would be the two KE's added together.

One needs to be very careful in using the power = change of momentum equation. It has to be dealt with for each individual particle, not net momentum change (else it would mean no KE change is ever possible within a discrete system!!!). Imagine the explosion in a sealed box - plenty of KE inside the box, but no change of momentum overall. One would have to write something like ∑√(Fi.vi)² for i=1 to n particles with different velocities (note, velocities are vectors, so particles can be only grouped into an 'n-th' ensemble particle if they have the same magnitude and direction).
vu is defined from a single inertial reference frame, that is not acted on by the thruster. That frame doesn't change is the whole point. vu, however, changes as the thruster operates. where M*d(vu)/dt = -F. The total momentum never changes, just like your box example.

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.

But this does also imply that the power consumption of the rocket depends on it's relative motion wrt to vu (ie it does depend on v = vr-vu). Unfortunately the math formatting here is a bit limited or I might be inclined to write it out more formally.
Last edited by kcdodd on Sat Feb 09, 2013 10:18 pm, edited 1 time in total.
Carter

Post Reply