## Rigorous analysis thrust/propellant power, arbitrary frame.

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

### Rigorous analysis thrust/propellant power, arbitrary frame.

Take a test mass, M kg, travelling at velocity Q. It begins ejecting mass at the rate of m kg/s of material at Vm/s at t=0. Now consider the energy required for this ejection rate in a time sample of Δt at t=0.

Impulse Equations (Force = rate of change of momentum):
Force on ejectant = d(m[Q+V])/dt = +V.m.Δt
Force on test mass = d((Q-(Vm/M))xM)/dt = -V.m.Δt

Conservation of momentum:
V.mΔt = -M.ΔQ

(note; ΔQ/Δt =~dQ/dt=-V.m/M, therefore Force (on test mass) = -mV (unsurprisingly, same result as above))

OK! So now do the maths!...

A: Power, by rate of change of kinetic energy, at t=0;
Test Mass---
ΔKE = ½ M Q² - ½ M (Q-ΔQ)² = +M.Q.ΔQ - ½ M ΔQ²
lim ΔQ->0; ΔQ²=0
So ΔKE = M.Q.ΔQ = -M.Q.mVΔt/M = -Q.V.m.Δt
Propellant---
ΔKE = ½ {(Q+V)² – Q²} m Δt = Q.V.m.Δt + ½ V².m.Δt

TOTAL ΔKE = -Q.V.m.Δt + Q.V.m.Δt + ½ V².m.Δt = ½ V².m.Δt

NOTE: TOTAL ΔKE independent of Q

B: Alternative calculation of Power, by rate of expended work, at t=0;
Test Mass---
Rate of Work = -V.m.(Q- ΔQ).Δt = -Q.V.m.Δt + Q.V.m.Δt.ΔQ
lim Δt->0; ΔQ.Δt=0
So Work rate = -Q.V.m.Δt
Propellant---
***CAUTION!!.. distance travelled by a mass accelerating to V in time Δt is V.Δt/2, because this is the average distance travelled for the linear acceleration (this is a legitimate assumption on a timescale of lim Δt->0)***
Rate of Work = V.m.(Q+V/2).Δt = Q.V.m.Δt + ½ .V².m.Δt

TOTAL rate of work = -Q.V.m.Δt + Q.V.m.Δt + ½ V².m.Δt = ½ V².m.Δt

TWO DIFFERENT APPROACHES; BY RATE OF WORK DONE AND BY ΔKE.

There is NO 'over-unity' violation if the maths is done correctly for an accelerated mass using a propellant.

Note that the ±Q.V.m.Δt term in each of the propellant and mass cancel out. This will grow as Q grows, but the result remains frame-independent because the sum of the terms is zero. Always. And forever. Because it is the same, but opposite, term in both the propellant and thrusted mass.

It only becomes an 'un-physical' violation if there is no propellant to counterbalance the Q.V.m.Δt term.

dkfenger
Posts: 30
Joined: Thu Jul 07, 2011 9:55 pm
Location: Victoria, BC
Thank you for this. I didn't like my earlier result, and I now see where I screwed up. Rounding 1.08N to 1.0N would only have been fine if I'd changed the propellant velocity to match. Without that, the KE calculation ends up unbalanced, and that's where the velocity dependence came from.

I really do know better to be careful when subtracting large values, but I got too caught up in the numbers to look at the math.

A simple derivation of KE being invariant to change in velocity:
Take two 1kg masses, initially at rest with respect to each other. Both are moving at velocity v m/s. Some time later, they are both moving at velocity v+dv m/s and v-dv m/s respectively. Ignore how that happened, and just evaluate KE for the two states:

KE0 = 0.5*2kg*v^2.
KE1 = 0.5*1kg*(v+dv)^2 + 0.5*1kg*(v-dv)^2
= 0.5*1kg*((v^2 + 2v*dv + dv^2) + (v^2 - 2v*dv + dv^2))
= 0.5*1kg*(2*v^2 + 2*dv^2)
= 0.5*2kg*v^2 + 0.5*2kg*dv^2

KE1-KE0 = 0.5*2kg*dv^2, independent of v, exactly as stated by chrismb above.

kcdodd
Posts: 722
Joined: Tue Jun 03, 2008 3:36 am
Location: Austin, TX
So, you're looking at how time rate of change of kinetic energy transforms for the rocket and propellant? Here's another.

KE = 0.5*m1*v1^2 + 0.5*m2*v2^2

take time derivative of KE.

dKE/dt = m1*v1*dv1/dt + m2*v2*dv2/dt

But, m1*dv1/dt = F1, and m2*dv2/dt = F2, the force on the two objects.

dKE/dt = v1*F1 + v2*F2

Now, we assume that F1 = -F2 from conservation of momentum.

dKE/dt = F1*(v1 - v2)

Ok, so now you can apply whatever transformation you want. Gallilean transformation is v1 = v1' - v, v2 = v2' - v. Plug in:

dKE/dt = F1*(v1' - v2')

independent of v.
Carter

CaptainBeowulf
Posts: 498
Joined: Sat Nov 07, 2009 12:35 am
Ok, well, I'll ignore secondhand accounts about what some of the physicists say about having to do Lorentz transforms and try to engage with this mathematical model for one or two posts.
It only becomes an 'un-physical' violation if there is no propellant to counterbalance the Q.V.m.Δt term.
This of course is true. You have to push against something in order to move. The theory behind the M-E thruster is, as best I remember it right now, that you push against distant matter in the universe by gravity waves reflected back to you.

So, if M-E theory is correct distant matter is the reaction mass.

Is there any way that as the M-E ship accelerates, Q would grow for it, and Q on the distant matter would grow too, resulting in the two summing to zero?

Is there an alternative mathematical description that makes sense for such a hypothetical M-E thruster? Or do you get to the same over-unity situation that GoatGuy did for years?

CaptainBeowulf
Posts: 498
Joined: Sat Nov 07, 2009 12:35 am
Note: Of course if you were just observing a M-E thruster accelerate without ejecting any propellant it would appear to reach the point where it would go over-unity, because you wouldn't know what it was pushing against. There would be no conservation of momentum.

I ask about an alternative mathematical model because there is, for example, no "m kg/s of material" being ejected - but there should (according to the theory) be force exerted on the distant matter. Presumably if you have a small, say 1kg M-E vehicle, and it's engaging with the bulk of the distant mass in the universe (quadrillions of kg? more?), the acceleration imparted to of all that distant mass would be pretty much undetectable (assuming you could somehow detect what was happening to huge amounts of distant mass) - but there should still be a force on it.