# How much power is "a power unit"

#### Lurking Grue

##### Banded Mongoose
(Off-topic, sorry.)

I read several of your blog entries (about 2300AD at blogspot) and must salute your attention to detail and striving for coherency & consistency for 2300AD. This is especially relevant and important in a game universe which aims for hard SF, like 2300AD does. The original 2300AD (i.e. Traveller 2300) was my first foray into hard SF, really, and holds a very special place in my heart (along with the 2nd edition). So it brings me a lot of joy seeing people who are still very dedicated to that game universe. Kudos!

#### Bill Sheil

##### Banded Mongoose
In Classic Traveller/Striker one displacement ton at TL15 produced 1 (Classic Traveller) "Power Unit" in starships which was 250MW in Striker.

In Mongoose 2 one DTon (TL15) produces 20 "(Mongoose) Power Units" so using the same paradigm one Mongoose "Power Unit" would be 12.5 MW.

Edit: Just realised you were talking KW rather then MW so your estimate and original Traveller differs by an order of 500 rather than 0.5. However, the Striker power ratings always seemed excessive and I believe that the original 2300AD although using a different design model substantially reduced the power levels from classic traveller.

By my quick calculations I think in a perfect lossless case ignoring reaction mass it would take 0.5-1MW power to generate 1G acceleration on a 1 DTon ship (10-20 ton mass) so assuming that 1 power plant DTon element (20 Power units) feeds 1 MDrive element to produce 1G thrust then the 1 Power unit would correspond to 25-50 KW.

Again this is with perfect efficiency and until someone with better physics comes along and corrects me.

Edit 2: Also the MDrive needs 10% of the tonnage in power per thrust which is half the points generate by an equivalent TL15 power plant unit so that makes the energy requirement Power Unit equivalent 50-100 KW.

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#### Geir

##### Cosmic Mongoose
In Classic Traveller/Striker one displacement ton at TL15 produced 1 (Classic Traveller) "Power Unit" in starships which was 250MW in Striker.

In Mongoose 2 one DTon (TL15) produces 20 "(Mongoose) Power Units" so using the same paradigm one Mongoose "Power Unit" would be 12.5 MW.
Based mainly on a liberal (literal? approximate? something) interpretation of the above, I've been assuming 10MW per power point, which is close. And don't look under the covers (or behind the curtain) and ask where all the waste heat is going....

#### Bill Sheil

##### Banded Mongoose
A key feature of the original traveller universe was the availability of cheap excessive energy but I seem to remember that the original GDW (Traveller) 2300AD did posit a substantially lower energy power plants so the two design paradigms were not directly comparable. 2300AD did not use power plant energy to directly generate thrust, they were used for weapons, stutterwarp and secondary systems. Mongoose is now sharing design sequences across both systems but it is not necessarily the case that power generation in both universes is the same.

#### Bill Sheil

##### Banded Mongoose
And don't look under the covers (or behind the curtain) and ask where all the waste heat is going....
Indeed this is the problem with the "big power" paradigm. The Expanse book/TV series had the same problem with excessive energy needed for their constant acceleration reaction mass drives.

##### Banded Mongoose
Traveller thruster plates, according to FF&S generate 40 tons of thrust per MW, or the power for 1 G is 0.25 MW per dTon. A 2 G 100 dTon Scout/Courier used 50 MW of power from the type A PP. This would scale a CT power point to 50 MW.

CT power plants are 1+Power*3 dTons. So 3 dTons of reactor would by 1 PP or (from above) 50 MW. Advanced fusion is 15 power per dTon. This solves close to 1 MW per power.

Reactionless drives are rather tricky to figure out in reality...

However, in 2k3 we have fission plants (the size of which we know from real subs etc.), conventional fuels (with a known energy content) and solar (with a known solar flux at Earth as standard). These resolve at close to 25 kW per power. Traveller fusion plants are a relative unknown...

#### Bill Sheil

##### Banded Mongoose
Traveller thruster plates, according to FF&S generate 40 tons of thrust per MW, or the power for 1 G is 0.25 MW per dTon. A 2 G 100 dTon Scout/Courier used 50 MW of power from the type A PP. This would scale a CT power point to 50 MW.

CT power plants are 1+Power*3 dTons. So 3 dTons of reactor would by 1 PP or (from above) 50 MW. Advanced fusion is 15 power per dTon. This solves close to 1 MW per power.

Thruster plates as an "alternative" technology for thrust in TNE terms does not really set a baseline for actual power generation and in any case in TNE FF&S TL15 power plants explicitly produce 6 MW per cubic metre - therefore about 80 MW per DTon at TL15 (or 3 DTons at TL12) so 1/20 would be 4MW for a Mongoose Power Unit.

CT power units are 1 per 3 DTons at TL12 but this is inconsistent with MgT2. In Classic Traveller TL15 power plants (1 power per DTon) were 3x better than the TL12 version (1 per 3 DTs) and the same for TNE, but in Mongoose the TL12 version generates 15 Power Units whereas the TL15 version generates 20 Power Units - only 33% better. But in any case CT Striker explicitly nails the CT Power Unit at 250 MW and translating the TL15 "baseline" between versions we get the 12.5 MW figure in Mongoose terms.

#### AnotherDilbert

##### Cosmic Mongoose
By my quick calculations I think in a perfect lossless case ignoring reaction mass it would take 0.5-1MW power to generate 1G acceleration on a 1 DTon ship (10-20 ton mass) so assuming that 1 power plant DTon element (20 Power units) feeds 1 MDrive element to produce 1G thrust then the 1 Power unit would correspond to 25-50 KW.

Again this is with perfect efficiency and until someone with better physics comes along and corrects me.
It's not quite that simple with constant acceleration...

Starting at rest at position 0, for simplicity:
Acceleration [m/s²] = a is constant,
Velocity [m/s] v = at,
Distance [m] d = at²/2,

Kinetic energy [J] E = mv²/2 = m(at)²/2 = ma²t²/2,
Power [J/s] is the time derivate of energy d/dt( ma²t²/2 ) = ma²t.
To maintain a constant acceleration requires an amount of power that increases linearly with time.

Example: Accelerate a 1 Dton ≈ 10 tonne craft by 1 G ≈ 10 m/s² for an hour = 3600 s:
Velocity achieved is v = at = 10 m/s² × 3600 s = 36000 m/s.
E = mv²/2 = 10000 kg × (36000 m/s)² / 2 = 6 480 000 000 000 J = 6480 GJ.
Average power is 6480 GJ / 3600 s = 1.8 GJ/s = 1.8 GW = 1800 MW = 1 800 000 kW.

Accelerate for two hours and the result is:
Velocity achieved is v = at = 10 m/s² × 7200 s = 72000 m/s.
E = mv²/2 = 10000 kg × (72000 m/s)² / 2 = 25 920 000 000 000 J = 25 920 GJ.
Average power is 25920 GJ / 7200 s = 3.6 GJ/s = 3.6 GW = 3600 MW = 3 600 000 kW.

Conclusion: The power consumed by the M-drive does not generate the acceleration of the craft directly, presumably it is used to trick the outside of the craft (system) to accelerate the craft. Hence the achieved acceleration cannot be used to estimate the power required according to our current technology.

Constant acceleration is just a vast simplification to avoid differential equations (such as the rocket equation) while playing Traveller.

See e.g.:
https://yourquickinfo.com/does-constant-power-mean-constant-acceleration/

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#### AnotherDilbert

##### Cosmic Mongoose
Traveller thruster plates, according to FF&S generate 40 tons of thrust per MW, or the power for 1 G is 0.25 MW per dTon. A 2 G 100 dTon Scout/Courier used 50 MW of power from the type A PP. This would scale a CT power point to 50 MW.
A Scout (PP A) produces 2 CT EP = 500 MW.
Scout: 100 Dt × 2 G = 2 EP
Free Trader: 200 Dt × 1 G = 2 EP

FFS scales power differently from CT/MT. A basic Scout-like ship needs ~500 MW in CT/MT, but only ~130 MW in FFS (roughly 85 MW propulsion + 10 MW contra-grav + 10 MW internal grav + 25 MW avionics).
Since the ship does not need massive amounts of HEPlaR fuel it's quite heavy and has to calculate acceleration by mass, not volume (the default short-cut).

Something like (TL-12):

#### Condottiere

##### Emperor Mongoose
It's a nebulous number.

By default, ten tonnes of hull requires one power point to accelerate it one gravity (Terran norm).

Apparently, one power point is required to energize the artificial gravity plates within that ten tonne hull, charitable a half point.

In the same vein, half a point is the minimum that all the lighting and appliances work in a ten tonne hull.

Four of which would power a beam laser to medium range.

An if you customize the manoeuvre drive, you only need a quarter power point to accelerate a ten tonne hull one gee.

Again, customization would allow a quarter power point to punch through dimensions and push through ten tonnes about three light years in about a week.

#### Bill Sheil

##### Banded Mongoose
It's not quite that simple with constant acceleration...

Starting at rest at position 0, for simplicity:
[snip]
Kinetic energy [J] E = mv²/2 = m(at)²/2 = ma²t²/2,
Power [J/s] is the time derivate of energy d/dt( ma²t²/2 ) = ma²t.
To maintain a constant acceleration requires an amount of power that increases linearly with time.

Example: Accelerate a 1 Dton ≈ 10 tonne craft by 1 G ≈ 10 m/s² for an hour = 3600 s:
[snip]
Average power is 6480 GJ / 3600 s = 1.8 GJ/s = 1.8 GW = 1800 MW = 1 800 000 kW.

Accelerate for two hours and the result is:
[snip]
Average power is 25920 GJ / 7200 s = 3.6 GJ/s = 3.6 GW = 3600 MW = 3 600 000 kW.

Conclusion: The power consumed by the M-drive does not generate the acceleration of the craft directly, presumably it is used to trick the outside of the craft (system) to accelerate the craft. Hence the achieved acceleration cannot be used to estimate the power required according to our current technology.

Constant acceleration is just a vast simplification to avoid differential equations (such as the rocket equation) while playing Traveller.
You were partly correct at least up to the "It's not quite that simple..." part but thereafter you are broadly wrong (with disclaimers).

The first and most basic issue is that you equate the (power plant) energy used to accelerate the mass with the "kinetic energy" of the final body. I specifically avoided referring to kinetic energy in my own calculation (which I really intended as just a ballpark figure) for this reason. In fairness the page you linked too leads on this as well without qualifying the basis or constraints for that assertion. It is a really bad page to be quite honest.

Kinetic energy is dependent on the frame of reference so if we have for example a 1000kg object (craft) that we observe (1) in our current frame to be travelling at 10m/s then the apparent kinetic energy of the craft in our frame is 50 kJ (0.5x1000x10x10 using E(kinetic)=0.5mv^2). If it accelerates (with its own power plant and negligible mass loss) to 20m/s then its apparent kinetic energy changes to 200 kJ. You assume that this required the input from its power plant of 150kJ.

But if we happen to be observing it from another craft travelling in parallel also at 10m/s then, in that frame of reference (2), the craft starts at zero velocity and zero kinetic energy then accelerates to 10 m/s... gaining 50 kJ kinetic energy.

Or if we observed from another equally valid frame of reference (3) travelling at a velocity of 1000m/s (E=500 MJ) relative to the craft's initial velocity and see the craft accelerate to a velocity of 1010 m/s (510.05 MJ) gaining over 10 MJ of kinetic energy.

Or (4) we started at 20 m/s in the same direction and saw the craft's accelerate from -10 m/s (50 kJ) to 0 m/s (0 J), in this case losing 50 kJ of energy.

So give that those four frames of reference are equally valid in classical and relative physics then how much power did the craft's power plant generate - (1) 150 kJ or (2) 50 kJ or (3) 10.05 MJ or (4) it absorbed 50 kJ?

The reality is that we cannot decouple the effect of acceleration/force on an object without considering the force on the reaction mass against which it is pushing (either ejected from the craft like a rocket or the system/planet that a thruster plate or bicycle is pushing against) this what actually levels this apparent inconsistency in energy. There is always another mass being moved and regardless of the frame of reference the same amount of power plant energy is being distributed between the "primary" craft and the reaction mass - just in different proportions dependent on the frame of reference.

In other word in a real system only a portion of the impulse energy is manifesting as a change of kinetic energy of the primary craft (could be over 100% but that is going down the rabbit hole) and likewise this is one of the reasons why reactionless systems are impossible.

I cannot continue tonight (work, and this is quite high demand) but I will say that in the case of Traveller thruster plates there is a degree to which constant acceleration will decouple the craft and its reaction mass so that the power requirement increases as the two masses accelerate away from one another, but this is further complicated by the motion of the craft through a system where it may couple to other reaction masses with other velocity vectors. So yes thruster plates could be even more complex.

#### Sigtrygg

##### Emperor Mongoose
Use Lagrangian mechanics rather than Newtonian and it makes a bit more sense.

#### AnotherDilbert

##### Cosmic Mongoose
You were partly correct at least up to the "It's not quite that simple..." part but thereafter you are broadly wrong (with disclaimers).
I'm glad we agree on something.

I hope we can agree on the simplistic maths in my post, even if you think I misapplied it to the physics at hand?

The first and most basic issue is that you equate the (power plant) energy used to accelerate the mass with the "kinetic energy" of the final body. I specifically avoided referring to kinetic energy in my own calculation (which I really intended as just a ballpark figure) for this reason. In fairness the page you linked too leads on this as well without qualifying the basis or constraints for that assertion. It is a really bad page to be quite honest.
I have no argument with ballpark figures. The linked page was just the first I found that seemed to discuss vaguely this problem. It was not intended as an insult to your intelligence.

Kinetic energy is dependent on the frame of reference so if we have for example a 1000kg object (craft) that we observe (1) in our current frame to be travelling at 10m/s then the apparent kinetic energy of the craft in our frame is 50 kJ (0.5x1000x10x10 using E(kinetic)=0.5mv^2). If it accelerates (with its own power plant and negligible mass loss) to 20m/s then its apparent kinetic energy changes to 200 kJ. You assume that this required the input from its power plant of 150kJ.

But if we happen to be observing it from another craft travelling in parallel also at 10m/s then, in that frame of reference (2), the craft starts at zero velocity and zero kinetic energy then accelerates to 10 m/s... gaining 50 kJ kinetic energy.

Or if we observed from another equally valid frame of reference (3) travelling at a velocity of 1000m/s (E=500 MJ) relative to the craft's initial velocity and see the craft accelerate to a velocity of 1010 m/s (510.05 MJ) gaining over 10 MJ of kinetic energy.

Or (4) we started at 20 m/s in the same direction and saw the craft's accelerate from -10 m/s (50 kJ) to 0 m/s (0 J), in this case losing 50 kJ of energy.

So give that those four frames of reference are equally valid in classical and relative physics then how much power did the craft's power plant generate - (1) 150 kJ or (2) 50 kJ or (3) 10.05 MJ or (4) it absorbed 50 kJ?
Agreed, there is no absolute frame of reference, but regardless of initial boundary conditions the rate of change of kinetic energy changes with time if acceleration is constant?

The reality is that we cannot decouple the effect of acceleration/force on an object without considering the force on the reaction mass against which it is pushing (either ejected from the craft like a rocket or the system/planet that a thruster plate or bicycle is pushing against) this what actually levels this apparent inconsistency in energy. There is always another mass being moved and regardless of the frame of reference the same amount of power plant energy is being distributed between the "primary" craft and the reaction mass - just in different proportions dependent on the frame of reference.
Agreed, we have to conserve both energy and momentum of the entire system, if indeed that is how M-drives work (I agree that is how it has to work, but it is not specified in Traveller as far as I know).

I don't understand how energy added to the system is applied directly to momentum (≈velocity) and the energy state is irrelevant.

If I solve the equation system momentum and energy (Ms = ship's mass, Vp = planet's velocity, a.s.o.)
MsVs + MpVp = 0
MsVs² + MpVp² + e = 0
and approximate away the term Ms²/Mp ≈ 0 as the planet's (solar system's?) mass is vastly larger than the ship's mass in general, all I get is the trivial
MsVs² ≈ e
MpVp² ≈ eMs/Mp² ≈ 0

As at least I would expect both objects gain the same momentum (in opposite directions), the smaller object gain a lot more velocity, hence the lion's share of the added energy?

For the same reason ion drives are painfully energy inefficient?

As you can see I approach this simplistically, from first order principles. I do not see, or understand, what I'm missing.

Are there some additional assumptions I have missed?

In other word in a real system only a portion of the impulse energy is manifesting as a change of kinetic energy of the primary craft (could be over 100% but that is going down the rabbit hole) and likewise this is one of the reasons why reactionless systems are impossible.

I cannot continue tonight (work, and this is quite high demand) but I will say that in the case of Traveller thruster plates there is a degree to which constant acceleration will decouple the craft and its reaction mass so that the power requirement increases as the two masses accelerate away from one another, but this is further complicated by the motion of the craft through a system where it may couple to other reaction masses with other velocity vectors. So yes thruster plates could be even more complex.

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#### AnotherDilbert

##### Cosmic Mongoose
Use Lagrangian mechanics rather than Newtonian and it makes a bit more sense.
Lagrange mechanics is about solving constrained systems exchanging kinetic and potential energy, such as pendulums or orbital mechanics, in the energy domain rather than the motion domain, if I understand it correctly.

As the discussed problem isn't constrained, I don't quite see how that would help?

If we get away a bit from the planet, the potential energy change involved would be negligible for Traveller ships with massive acceleration?

#### Condottiere

##### Emperor Mongoose
If a manoeuvre drive was just the creation of a gravitational force that allows the hull to push through space, you wouldn't need to anchor it to a (large) gravitational well.

Doing that, sounds more like it taps into existing forces, possibly with strings attached.

Well, that would be my theory.

#### Sigtrygg

##### Emperor Mongoose
Lagrange mechanics is about solving constrained systems exchanging kinetic and potential energy, such as pendulums or orbital mechanics, in the energy domain rather than the motion domain, if I understand it correctly.

As the discussed problem isn't constrained, I don't quite see how that would help?

If we get away a bit from the planet, the potential energy change involved would be negligible for Traveller ships with massive acceleration?
No, you can use Lagrangians for any classical physics necessity.
This is a nice introduction:

#### AnotherDilbert

##### Cosmic Mongoose
No, you can use Lagrangians for any classical physics necessity.
This is a nice introduction:
Isn't that exactly what I did in my first calculation?

I choose constraints and generalised coordinate(s) (only t):
Starting at rest at position 0, for simplicity:
Acceleration [m/s²] = a is constant,
Velocity [m/s] v = at,

Defined the Lagrangian function:
Kinetic energy [J] E = mv²/2 = m(at)²/2 = ma²t²/2,

That is trivial to integrate from t=0 to 3600 s:
Example: Accelerate a 1 Dton ≈ 10 tonne craft by 1 G ≈ 10 m/s² for an hour = 3600 s:

Velocity achieved is v = at = 10 m/s² × 3600 s = 36000 m/s.
E = mv²/2 = 10000 kg × (36000 m/s)² / 2 = 6 480 000 000 000 J = 6480 GJ.

As I have already chosen only one generalised coordinate (t), the Euler-Lagrange equation system collapses?
Power [J/s] is the time derivate of energy d/dt( ma²t²/2 ) = ma²t.
That is not constant, but a function of time?

#### Bill Sheil

##### Banded Mongoose
Agreed, there is no absolute frame of reference, but regardless of initial boundary conditions the rate of change of kinetic energy changes with time if acceleration is constant?
Which is irrelevant because the mistake you are still making is assuming that the kinetic energy change of the craft is somewhat proportional to the energy that is provided by the power plant. They are not directly linked.

See my example. And note it was only a single example, same craft in the same place with the same vector in the same universe doing the same acceleration. All my example does is view this from a number of different frames of reference, there are no variations in "initial boundary conditions" but the change in kinetic energy calculated from each of those equally valid frames of reference is radically different: 50 KJ, 150 KJ, 10 MJ, -50 KJ. All of these are the same example just measured from a different viewpoint.

And also there is no reason why the frame of reference needs to be constant. From the frame of the craft's passengers the acceleration is always "from zero" instant by instant.

My point is that you cannot say : "the power plant generates 50 KJ of energy (after losses) to the thruster and the craft gains 50 KJ of Kinetic energy". The change in kinetic energy is different in different frames of reference. There is always one frame of reference where the craft does gain 50 KJ of KE but there is another equally valid frame of reference where the craft loses 10 GJ of kinetic energy when you give its thruster 50 KJ of energy. The energy provided by a power plant is invariant (in a Galilean sense) but the Kinetic Energy is not.

For the same reason you cannot generally say "it takes more power plant energy to accelerate at a fixed rate at high velocities because the change in kinetic energy is greater" (see later for Thruster plates). Power plant energy (invariant) and kinetic energy (frame of reference variant) are not the same. The higher/increasing velocity in one frame of reference could be is a lower/decreasing velocity in many other frames of reference.
For example let's take your original example:
Example: Accelerate a 1 Dton ≈ 10 tonne craft by 1 G ≈ 10 m/s² for an hour = 3600 s:
Velocity achieved is v = at = 10 m/s² × 3600 s = 36000 m/s.
E = mv²/2 = 10000 kg × (36000 m/s)² / 2 = 6 480 000 000 000 J = 6480 GJ.
Average power is 6480 GJ / 3600 s = 1.8 GJ/s = 1.8 GW = 1800 MW = 1 800 000 kW.

Accelerate for two hours and the result is:
Velocity achieved is v = at = 10 m/s² × 7200 s = 72000 m/s.
E = mv²/2 = 10000 kg × (72000 m/s)² / 2 = 25 920 000 000 000 J = 25 920 GJ.
Average power is 25920 GJ / 7200 s = 3.6 GJ/s = 3.6 GW = 3600 MW = 3 600 000 kW.

But we will start at a frame of reference with a vector of 50000m/s in the direction of acceleration without actually changing anything real.
From this viewpoint the 10 tonne craft starts with a velocity of -50000 m/s, kinetic energy of 12500 GJ.
After one hour of acceleration it has reduced its velocity to 14000m/s, kinetic energy: 980 GJ (reduced by 11520 GJ), average Power -3.2 GJ (minus).
After two hours the craft is travelling at 22000 m/s in our frame of reference. Kinetic energy is 2420 GJ (10 GJ less than the starting KE and at one point it actually dropped to zero before rising again).

Nothing real has changed in this example from your earlier calculations but your general supposition that power plant energy increases (feeds into) kinetic energy is false. The power plant has not transmuted into an energy sink in this example. It is the same example as you provided earlier when you calculated that the kinetic energy would increase. I made no changes.

The crux of this is that we always need to consider the whole mass system not just the primary (craft) but the reaction mass as well. The reaction mass has its own, mass, velocity, momentum and kinetic energy. When we do this the invariant Power Plant energy is distributed between both the primary mass and the reaction mass (both receive equal and opposing momentum). The total change in kinetic energy of both the primary and the reaction mass equals the total propulsion (power plant) energy but the proportional distribution depends on the frame of reference.

When we have a frame of reference where the kinetic energy seems to be is reducing it because in that frame of reference the reaction mass is gaining more energy than the power plant is providing (the combined primary loss and reaction excess cancel out).

Conclusion: The power consumed by the M-drive does not generate the acceleration of the craft directly, presumably it is used to trick the outside of the craft (system) to accelerate the craft. Hence the achieved acceleration cannot be used to estimate the power required according to our current technology.

Again not necessarily true - the primary craft can gain more kinetic energy that the power plant provides so long as this is balanced by the reaction mass losing kinetic energy - e.g when a hurtling rocket ejects a reaction mass at an ejection velocity close to or slower than its travel velocity in a specific frame - so the ejected reaction mass seems to have reduced velocity and has lost kinetic energy relative to its velocity/KE when it was a component of the rocket mass.

But I want to circle back. In Traveller there are two basic drive models: Reaction mass drives and Thruster Plates.

Reaction Mass Drives (CT Fusion drives, HEPlaR) emit mass carried from the ship so they constitute a closed system. Propulsion energy basically goes into accelerating and ejecting the reaction mass and this can be modelled from an accelerating frame of reference anchored to the rocket. If the propulsion energy put into the reaction mass is constant for the same mass emission rate then its acceleration, exit velocity, kinetic energy are constant and the momentum it generates for the craft's acceleration is also constant, regardless of the external frame of reference or apparent velocity in any classic frame of reference at least until the rocket equation starts to kick in.

Thruster Plates (a.k.a Space Bicycle Drive) pushes against nearby Planetary/Stellar objects through some field. The local Planetary System in this case is the reaction mass. In this case the relative velocity of the craft and the reaction mass (i.e the average motion of the surrounding environment) does impact the kinetic energy and the propulsion input will have an impact on velocity generated. If the ship and the reaction mass have a positive relative velocity (ship generally leaving the system) then the acceleration will reduce as the relative velocity increases. If the ship is moving into the system mass then the acceleration per power provided should increase. This is also complicated by the localisation of the field which may be decoupling from some mass as it moves away and finding new mass as the craft moves through the system. Also if the thruster tends to decouple from mass that has a large relative velocity as it accelerates (it cannot "grip" it) it may also recouple to nearby mass that has a similar velocity to the craft's vector and therefore gives it more grip.

There was also an abomination of a reactionless thrust mechanism in MegaTraveller but I am not going to address that and all its infinite energy/perpetual motion implications.

#### AnotherDilbert

##### Cosmic Mongoose
Which is irrelevant because the mistake you are still making is assuming that the kinetic energy change of the craft is somewhat proportional to the energy that is provided by the power plant. They are not directly linked.

See my example. And note it was only a single example, same craft in the same place with the same vector in the same universe doing the same acceleration. All my example does is view this from a number of different frames of reference, there are no variations in "initial boundary conditions" but the change in kinetic energy calculated from each of those equally valid frames of reference is radically different: 50 KJ, 150 KJ, 10 MJ, -50 KJ. All of these are the same example just measured from a different viewpoint.

And also there is no reason why the frame of reference needs to be constant. From the frame of the craft's passengers the acceleration is always "from zero" instant by instant.
The energy gained (or lost) by the ship is different in different frames of reference, but total energy is still conserved in each frame of reference.

The ship and the reaction mass forms a closed system that conserves momentum and energy. If the ship and the reaction mass starts accelerating away from each other the total energy of the system increases, that energy has to come from somewhere and the power required to maintain constant acceleration will increase with time. I don't see where the power comes from, if not the drives of the ship?

In the frame of the ship, the reaction mass keeps accelerating away from the ship, it does not start from zero in every instant.

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