Mach Effect progress

[Barry Kirk]

In terms of the energy and thrust balance issues. If one looks at a thruster moving at less than 0.1 c... Could you use standard Newtonian mechanics to understand what is going on.

What I'm saying is that you put the thruster inside a black box and disregard anything happening inside the box. Just what is going into and out of the box.

Can you use Newtonian mechanics to describe the energy of the box?

[chrismb]

He was doing just as GoatGuy, Andrew and chris are doing and failing to note that thrust or better "force" is only constant when the thruster is stationary.

chrismb has NOT DONE [THUS CANNOT HAVE FAILED TO DO] any calculations.

chrismb posts are asking for this supposed violation of KE regarding conventional thrusters to be shown.

Forget 'ME theory' for the moment. Just show what GIT claims is a demonstration of how the maths wrongly shows a violation of KE with conventional rockets.

JUST SHOW UP THE EQUATIONS!

[GIThruster]

I'm not the one objecting chris. You show the calculations or spare us your pathetic, childish temper tantrums.

[GIThruster]

What I'm saying is that you put the thruster inside a black box and disregard anything happening inside the box. Just what is going into and out of the box.

Can you use Newtonian mechanics to describe the energy of the box?.

No. The conservation issue does not arise unless you allow the thruster to accelerate. Once you do that, using newtonian mechanics you will necessarily show a conservation violation, regardless of the kind of thruster you use.

[chrismb]

Once you do that, using newtonian mechanics you will necessarily show a conservation violation, regardless of the kind of thruster you use.

Show this violation.

[dkfenger]

OK, here's some calculations. I dug up the datasheet for Busek high-power hall effect engines:
Busek BHT-20k Hall Effect Thruster
Discharge Input Power: 20kW
Discharge Voltage: 500V
Discharge Current: 40.5A
Propellant Mass Flowrate: 40.0 mg/sec
Thrust: 1.08N
Specific Impulse: 2750 sec
Propulsive Efficiency: 70%


Let's build a hypothetical microsatellite around this with a handwavium on-board power source.
10kg craft, 4kg is propellant. Round the numbers, thrust is 1N at 40mg/sec.

Exhaust velocity = 9.8066 m/s^2 * 2750s = 26,968.15 m/s. Call it 27km/s for simplicity.
Back-check delta-p: 40mg * 27km/s = 0.00004 kg * 27000m/s = 1.08 N s, so 40mg/sec produces 1.08N of thrust. Good.


Examine the effects of 1s of thrust (should not change frame enough to matter):
a = F/m = 1N/10kg = 0.1 m/s^2. Over 1s, change velocity of craft by 0.1m/s.
Energy dissipated: 20kJ.
Change in KE of craft:
KE0 = 0.5*v0^2*10kg
KE1 = 0.5*(v0+0.1m/s)^2*(10kg - 40mg)

Measure in different frames based on v0 of craft:
v0 KE0 KE1 delta-KE
0m/s 0J 0.0499998J 0.05J
1e4 m/s 5e8J 5.00008e8J 8kJ
2e4 m/s 2e9J 2.000012e9J 12kJ
4e4 m/s 8e9J 8.000008e9J 8kJ
5e4 m/s 1.25e10J 1.25e10J 0kJ

The hypothetical Hall Effect rocket above *never* goes over-unity.

Let's try again with a 1kg craft instead: delta-V is now 1m/s for 1s of thrust. delta-m is 40mg, still.
at 1e4 m/s delta-KE is 8kJ
at 2e4 m/s delta-KE is 12kJ
at 2.5e4 m/s delta-KE is 12.5kJ
at 3e4 m/s delta-KE is 12kJ and falling

Note that at the peak, the KE of the exhaust fuel is approximately zero. This shouldn't come as much of a surprise, but was something I wasn't familiar with.

KE of exhaust fuel when craft is at 0m/s: 0.5*(2.7e4m/s)^2*40mg = 14.58kJ
Plotting fuel+craft delta-KE produces a straight line, starting at 14.58kJ at v0=0m/s, hitting 0 at v0=1.82e5m/s. [Which suggests that if you fire the craft backward while at high velocity, the *propellant* violates energy conservation. Perhaps this negates my point, perhaps not. Or perhaps I've done this entirely wrong. I would welcome any corrections.]

Note 14.58kJ / 20kJ = about the 70% propulsive efficiency claimed on the datasheet.

Would anyone care to present the same math for a 10kg Mach Effect spacecraft? How much thrust could one expect from a 20kW Mach Effect thruster under realistic assumptions about the new ceramics, assuming the effect is real?

[GIThruster]

No chris. It has been illustrated by GoatGuy, Andrew and you, but you all make the same mistake. There is no onus upon me to do your math for you. Do it yourself and do it with a chemical thruster on an arm with sufficient fuel provided. You'll see the same conservation violation arises which should be no surprise because GRT is the correct scientific theory to explain such motion--not newtonian theory.

[GIThruster]

dkfenger, I just glanced at your calculation but just noting, you take propulsive efficiency as a constant or invariant, which is wrong, It is not. Your method cannot give the proper result. Thrust is not relativistically invariant.

[chrismb]

No chris. It has been illustrated by GoatGuy, Andrew and you, but you all make the same mistake.

Provide a link to a post in this forum where any of these people have illustrated this 'mistake'.

GIT is claiming something which he can simply close down by putting forward what he claims to say exists. It is a link, or copied text. How difficult is that?

[chrismb]

dkfenger, I just glanced at your calculation but just noting, you take propulsive efficiency as a constant or invariant, which is wrong, It is not. Your method cannot give the proper result. Thrust is not relativistically invariant.

Rubbish. This is basic school-level Newtonian mechanics. Einstein wasn't needed for this calculation. He's stated an assumption, a 'curly delta' velocity/mass at v=0. It's accurate enough for its purpose.

GIT, now show the forum the calculation which shows violation. C'mon, man!

[GIThruster]

chris, please explain to your audience why it is we should not do newtonian dynamics in non-inertial frames.

[dkfenger]

Expanding on the calculation for craft and propellant KE when the hall effect thruster is used to decelerate.

KE0 = 0.5*v0^2*1kg
KEc1 = 0.5*(v0-1m/s)^2*(1kg-40mg)
KEp1 = 0.5*(v0+2.7e4m/s)^2*40mg
KE1 = KEc1 + KEp1

v0 of 1e4 m/s => delta-KE of 1.5381e4 J
v0 of 2e4 m/s => delta-KE of 1.6181e4 J
v0 of 4e4 m/s => delta-KE of 1.7783e4 J
v0 of 8e4 m/s => delta-KE of 2.0984e4 J

That's higher than the 20kJ energy input for the 1s the thruster fires, and thus a violation of KE from the frame where the craft is moving at 8e4 m/s. In the craft's original frame, the change in KE is 14.57kJ.

[chrismb]

chris, please explain to your audience why it is we should not do newtonian dynamics in non-inertial frames.

dkfenger HAS stuck to an inertial frame, and calculated the 'instantaneous' [close enough, for an approximation] KE change for DIFFERENT, BUT INVARIANT inertial frames.

[dkfenger]

For purposes of argument, the t=0 and t=1s frames can be treated as two separate inertial frames. I was not trying to calculate the KE in the craft's frame, but in an inertial frame in which the craft's velocity at t=0 was specified.

I think I've actually served to prove GIT's argument more than chrismb's, however. delta-KE for a rocket changes based on the frame of reference. (Which I think was what GIT was getting at with talk of invariants.) Thus, the fact that there is a frame where the delta-KE of a hypothetical Mach Effect thruster is over-unity does not in and of itself violate physical law.

Still seems like cheating, though.

[GIThruster]

I'm exhausted. Jim was right to warn me about engaging people like this.

Yeah chris. You must be right. All the world of real physicists are wrong and you're right. Why don't you correct the world by trying to publish in a peer reviewed physics journal? Don't you want to show how completely clever you are? Show EVERYONE how clever chris is!

Sheesh. . .