# Sci-fi engine thrust calculations

• Deleted member 690984
In summary, a technical manual is being written for a starship in a science-fiction series. The ship's sub-light engines are capable of a maximum speed of 15,000 km/s (0.05c) and require 18,750 teraNewtons of thrust to achieve this speed at a fully loaded mass of 5 million metric tons. The ship also has 30 reaction control thrusters that can produce a combined thrust of 375,000 gees for manoeuvring. The ship's maximum speed is limited by practical considerations to prevent time dilation effects. The fully loaded ship mass includes all components.
Deleted member 690984
Hi all,

I'm writing a technical manual for a starship in a science-fiction series I'm working on, I wanted to check with some folks here to make sure my calculations are correct concerning the ship's engine thrust.

The ship has a fully-loaded mass of 5 million metric tons. Using its sub-light engines, it is capable of a maximum speed of 15,000 km/s (0.05c). Assuming 1 Newton can move 1kg of mass at 1 m/s per second, and given the ship has two sub-light engines, I've calculated it as requiring a thrust output of 18,750 teraNewtons of thrust to achieve 15,000 km/s at 5 million metric tons. Is this correct?

The ship also has several reaction control thrusters, 30 of them (5 fwd, 5 aft, 5 starboard, 5 port, 5 dorsal, and 5 ventral) for manoeuvring. Each thruster is actually a group of 5 exhausts, each of which can produce 5 megaNewtons of thrust (so 25 megaNewtons per thruster, essentially). I've calculated this as giving it at top possible speed using just these thrusters as 0.36 km/s (or 1,296 km/h). Is this correct?

Sorry, I've more questions than answers, @paulthomas, such as, what calculation did you use to derive your numbers above?

And I assume your sub-light engines expelling reaction mass in some fashion? Does this mean that the maximum speed is attained when the reaction mass is exhausted? Or is the ship bound by a practical maximum speed that is half the reaction mass to ensure it can stop at the other end of its trip?

Does your fully loaded ship mass include the fuel? How long does the ship take to reach its maximum speed? And what acceleration do the sub-light engines provide? Have you tried applying the rocket equation to your scenario?

As an aside, is the number of engines material, given that you've arbitrarily assigned two to this ship and nominated the maximum speed? And won't some of those reaction control thrusters just push the ship around in a circle?

Deleted member 690984
paulthomas said:
It doesn't take any thrust to maintain constant speed. Thrust is only needed for acceleration. See Newton's first and second laws of motion (which I assume hold in your fictitious universe).

With constant thrust there is no maximum speed, except ultimately the speed of light.

Deleted member 690984
paulthomas said:
The ship has a fully-loaded mass of 5 million metric tons. Using its sub-light engines, it is capable of a maximum speed of 15,000 km/s (0.05c). Assuming 1 Newton can move 1kg of mass at 1 m/s per second, and given the ship has two sub-light engines, I've calculated it as requiring a thrust output of 18,750 teraNewtons of thrust to achieve 15,000 km/s at 5 million metric tons. Is this correct?
##5 \times 10^{9}\ \text{kg}## mass. (5 million metric tons)
##1.875 \times 10^{16}\ \text{N}## thrust (18,750 teraNewtons) per engine

I have no idea how you get 15,000 km/s out of that. Oh. Maybe I see it. I think you divided total thrust by mass and then by 5 (for five million metric tons?) again. And then you slipped a decimal place or two.

The correct calculation is:

Multiply per-engine thrust by two to get total thrust. Divide by mass to get acceleration: ##3.75 \times 10^6\ \text{m}/\text{sec}^2##. This does not give you a velocity. It gives you an acceleration rate.

Divide by 10 (##10\ \text{m}/\text{sec}^2## = 1 gee) to get acceleration in gees. 375,000 gees. One hopes that your inertial dampers are good or the crew will be jelly on the floor.

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Deleted member 690984 and Melbourne Guy
Melbourne Guy said:
Sorry, I've more questions than answers, @paulthomas, such as, what calculation did you use to derive your numbers above?

I calculated it based on the fact that 1 Newton of force is required to push 1kg of mass at 1 meter per second. However, I think I may have misinterpreted this equation (I'm certainly not a physicist or a mathematician - I in fact have mild dyscalculia, hence why I'm here). I converted the mass of the ship into kilograms (so 5 billion kilos), and then calculated the speed I wanted it to go, and derived the Newtons from that.

Melbourne Guy said:
And I assume your sub-light engines expelling reaction mass in some fashion? Does this mean that the maximum speed is attained when the reaction mass is exhausted? Or is the ship bound by a practical maximum speed that is half the reaction mass to ensure it can stop at the other end of its trip?

Yes, the ship is bound by a practical maximum speed limit when using the sub-light engines to prevent time dilation. I chose 1/20 the speed of light as time dilation effects (correct me if I'm wrong) don't become too much of an issue until you approach about 1/10 the speed of light.

Melbourne Guy said:
Does your fully loaded ship mass include the fuel? How long does the ship take to reach its maximum speed? And what acceleration do the sub-light engines provide? Have you tried applying the rocket equation to your scenario?

As an aside, is the number of engines material, given that you've arbitrarily assigned two to this ship and nominated the maximum speed? And won't some of those reaction control thrusters just push the ship around in a circle?

Yes, the fully loaded ship mass includes everything - fuel, crew, cargo, etc. I haven't yet decided on a rate of acceleration for the ship. As for the RCS thrusters, not all of them are firing simultaneously. They only fire when the ship needs to manoeuvre.

PeroK said:
It doesn't take any thrust to maintain constant speed. Thrust is only needed for acceleration. See Newton's first and second laws of motion (which I assume hold in your fictitious universe).

With constant thrust there is no maximum speed, except ultimately the speed of light.
Of course - I chose 15,000 km/s as a practical maximum speed limit (of course in theory, as you say, the ship can travel as fast as it wants with sub-light engines, except at the speed of light) because it would minimise time dilation when traveling without warp engines.

paulthomas said:
Of course - I chose 15,000 km/s as a practical maximum speed limit
It doesn't alter the fact that any thrust (however small) will get you to any sub-light speed. And that thrust is about acceleration, not speed.

Deleted member 690984
jbriggs444 said:
##5 \times 10^{9}\ \text{kg}## mass. (5 million metric tons)
##1.875 \times 10^{16}\ \text{N}## thrust (18,750 teraNewtons) per engine

I have no idea how you get 15,000 km/s out of that. Oh. Maybe I see it. I think you divided total thrust by mass and then by 5 (for five million metric tons?) again. And then you slipped a decimal place or two.

The correct calculation is:

Multiply per-engine thrust by two to get total thrust. Divide by mass to get acceleration: ##3.75 \times 10^6\ \text{m}/\text{sec}^2##. This does not give you a velocity. It gives you an acceleration rate.

Divide by 10 (##10\ \text{m}/\text{sec}^2## = 1 gee) to get acceleration in gees. 375,000 gees. One hopes that your inertial dampers are good or the crew will be jelly on the floor.
Yeah, going off the other replies here, I think I've made a fundamental misunderstanding in the equation. I was thinking it gives you a rate of velocity, when in fact it's a rate of acceleration.

jbriggs444
PeroK said:
It doesn't alter the fact that any thrust (however small) will get you to any sub-light speed. And that thrust is about acceleration, not speed.
Gah. Yeah. I've misunderstood the equation entirely.

As a side, is there an equation I can use that would allow me to calculate how much fuel the sub-light engines would require, whatever their thrust is? Assuming the fuel is aneutronic fusion, using Helium-3 and deuterium.

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jbriggs444
I can't make heads nor tails of these equations or how to apply them. I'd appreciate if someone could help me out here.

I've finalised my ship's mass at 5,227,500 metric tons - however, that's it's loaded tonnage, including crew, food, fuel, cargo, etc. I've set a deadweight tonnage of 1,478,300 metric tons, so without any crew, food, fuel, cargo, etc. onboard, the ship has a mass of 3,749,200 metric tons.

The ship's secondary sub-light engines produce 130.7 teraNewtons of force each (2 aft, for accelerating, 2 forward, for decelerating), so 261.4 teraNewtons total for each set of 2 engines. I've calculated this as producing an acceleration of 50,000 m/s2, which would take 5 minutes to reach a speed of 15,000 km/s (0.05c) when the ship is fully loaded.

However, I just can't figure out how much energy that would require for the ship at loaded tonnage to go from 0 to 15,000 km/s in a 5 minute burn of the engines. Each engine is powered by a pair of aneutronic fusion reactors, which fuse Helium-3 and Deuterium and channels the plasma to the exhaust, where it is compressed and expelled by a powerful magnetic field to amplify the thrust.

PeroK
Yes, fiction is easier than mathematics!

PeroK said:
Yes, fiction is easier than mathematics!
Indeed! Are you able to help with my above problem at all?

paulthomas said:
Indeed! Are you able to help with my above problem at all?
It's a novel, not homework. Think artistically.

"Here the ship sailed out into the blue and sunny morn
The sweetest sight ever seen."

PeroK said:
It's a novel, not homework. Think artistically.

"Here the ship sailed out into the blue and sunny morn
The sweetest sight ever seen."
I want to make it as close to reality as I can. Can you help or not?

paulthomas said:
I want to make it as close to reality as I can. Can you help or not?
The reality is that we are earthbound.

PeroK said:
The reality is that we are earthbound.
So no, you can't, or do not want to. Please stop wasting my time.

paulthomas said:
So no, you can't, or do not want to. Please stop wasting my time.
I know nothing about aneutronic fusion reactors, so no I can't help!

PeroK said:
I know nothing about aneutronic fusion reactors, so no I can't help!
Then just say so to begin with, instead of being obtuse.

paulthomas said:
The ship's secondary sub-light engines produce 130.7 teraNewtons of force each (2 aft, for accelerating, 2 forward, for decelerating), so 261.4 teraNewtons total for each set of 2 engines. I've calculated this as producing an acceleration of 50,000 m/s2, which would take 5 minutes to reach a speed of 15,000 km/s (0.05c).

However, I just can't figure out how much energy that would require for the ship at loaded tonnage to go from 0 to 15,000 km/s in a 5 minute burn of the engines. Each engine is powered by a pair of aneutronic fusion reactors, which fuse Helium-3 and Deuterium and channels the plasma to the exhaust, where it is compressed and expelled by a powerful magnetic field to amplify the thrust.
Let us work with algebra.

We are not really in a relativistic range, so ##KE = 1/2 mv^2## to adequate precision.

Ideally, a rocket will expel its burnt fuel out the back as reaction mass with whatever velocity can be achieved by the energy from burning said fuel.

So our exhaust velocity ##v_e## can be computed from our energy density ##e## as ##e = 1/2 {v_e}^2##.

Our energy density is 18.3 MeV per 5 Daltons -- nuclear fusion of Helium-3 with Hydrogen-2 yielding Helium-4 plus a proton. Call this energy density ##e##.

Our desired thrust (130 teraNewtons) dictates our momentum flow rate. Call this F. A force is nothing other than a rate of change of momentum over time.

Our mass flow rate be the momentum flow rate divided by the exhaust velocity. Call this ##\dot{M}##.

Our energy flow rate will be the mass flow rate multiplied by the energy density. Call this ##P## for "power".

Now to the algebra. We want power P.

$$F = 130 \times 10^{12} \text{Newtons}$$
$$e = 18.3\ \text{MeV}/5 \text{Daltons} = 3.5 \times 10^{14}\ \text{joules}/\text{kg}$$
$$v_e = \sqrt{2e}$$
$$\dot{M} = \frac{F}{v_e} = \frac{F}{\sqrt{2e}}$$
$$P=e\dot{M} = e \frac{F}{v_e} = e \frac{F}{\sqrt{2e}} = F\frac{\sqrt{2e}}{2} = 130 \times 10^{12} \frac {\sqrt{2 \times 3.5 \times 10^{14}}} {2}$$

I make it ##1.7 \times 10^{21}## watts -- 1.7 zettawatts per engine.

Deleted member 690984
jbriggs444 said:
Let us work with algebra.

We are not really in a relativistic range, so ##KE = 1/2 mv^2## to adequate precision.

Ideally, a rocket will expel its burnt fuel out the back as reaction mass with whatever velocity can be achieved by the energy from burning said fuel.

So our exhaust velocity ##v_e## can be computed from our energy density ##e## as ##e = 1/2 {v_e}^2##.

Our energy density is 18.3 MeV per 5 Daltons -- nuclear fusion of Helium-3 with Hydrogen-2 yielding Helium-4 plus a proton. Call this energy density ##e##.

Our desired thrust (130 teraNewtons) dictates our momentum flow rate. Call this F. A force is nothing other than a rate of change of momentum over time.

Our mass flow rate be the momentum flow rate divided by the exhaust velocity. Call this ##\dot{M}##.

Our energy flow rate will be the mass flow rate multiplied by the energy density. Call this ##P## for "power".

Now to the algebra. We want power P.

$$F = 130 \times 10^{12} \text{Newtons}$$
$$e = 18.3\ \text{MeV}/5 \text{Daltons} = 3.5 \times 10^{14}\ \text{joules}/\text{kg}$$
$$v_e = \sqrt{2e}$$
$$\dot{M} = \frac{F}{v_e} = \frac{F}{\sqrt{2e}}$$
$$P=e\dot{M} = e \frac{F}{v_e} = e \frac{F}{\sqrt{2e}} = F\frac{\sqrt{2e}}{2} = 130 \times 10^{12} \frac {\sqrt{2 \times 3.5 \times 10^{14}}} {2}$$

I make it 1.7 \times 10^21 watts -- 1.7 zettawatts per engine.
Fantastic! And this is accounting for a 5 minute burn of the engines and the ship at loaded mass (just over 5.2 million metric tons)?

paulthomas said:
Fantastic! And this is accounting for a 5 minute burn of the engines and the ship at loaded mass (just over 5.2 million metric tons)?
A watt is a joule per second. 1.7 times 10^21 joules every second for the duration of the 300 second burn. On each engine.

You should probably look at the mass flow rate to make sure you do not run out.

Deleted member 690984
jbriggs444 said:
A watt is a joule per second. 1.7 times 10^21 joules every second for the duration of the 300 second burn. On each engine.

You should probably look at the mass flow rate to make sure you do not run out.
Just to make sure I have understood this right:

It requires 1.7 zettawatts per second, for 300 seconds (so 510 zettawatts), on each of the two engines facing aft, to achieve a 50,000 m/s2 acceleration, to push the ship to 15,000 km/s? So a total of 1,020 zettawatts?

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Sorry to be a pain, but assuming the above is correct, how much fuel would be required to achieve 1,020 zettawatts?

I've just thought of a new sci-fi invention in this regard - a Triple Alpha Reactor, one that would fuse deuterium and lithium-6 (22.4 MeV per fusion), which produces two helium-4 atoms. The reactor would then fuse those two helium-4 atoms together into carbon-12 (a net energy release of 7.275 MeV), so, effectively, 29.675 per fusion.

With this in mind, how much fuel would be required to achieve 1,020 zettawatts?

paulthomas said:
Sorry to be a pain, but assuming the above is correct, how much fuel would be required to achieve 1,020 zettawatts?

I've just thought of a new sci-fi invention in this regard - a Triple Alpha Reactor, one that would fuse deuterium and lithium-6 (22.4 MeV per fusion), which produces two helium-4 atoms. The reactor would then fuse those two helium-4 atoms together into carbon-12 (a net energy release of 7.275 MeV), so, effectively, 29.675 per fusion.

With this in mind, how much fuel would be required to achieve 1,020 zettawatts?
That's the mass flow rate. Let's do that one...
$$F = 130 \times 10^{12} \text{Newtons}$$
$$e = 18.3\ \text{MeV}/5 \text{Daltons} = 3.5 \times 10^{14}\ \text{joules}/\text{kg}$$
$$v_e = \sqrt{2e}$$
$$\dot{M} = \frac{F}{v_e} = \frac{F}{\sqrt{2e}} = \frac{130 \times 10^{12}}{\sqrt{2 \times 3.5 \times 10^{14}}}$$
I make that 4.9 million kg/sec for 300 seconds for a total of 1.4 million metric tons per engine.

Deleted member 690984
paulthomas said:
Sorry to be a pain, but assuming the above is correct, how much fuel would be required to achieve 1,020 zettawatts?

I've just thought of a new sci-fi invention in this regard - a Triple Alpha Reactor, one that would fuse deuterium and lithium-6 (22.4 MeV per fusion), which produces two helium-4 atoms. The reactor would then fuse those two helium-4 atoms together into carbon-12 (a net energy release of 7.275 MeV), so, effectively, 29.675 per fusion.

With this in mind, how much fuel would be required to achieve 1,020 zettawatts?
This new fuel is 22.4 Mev out of 8 Daltons worth of input.
The old fuel is 18.3 Mev out of 5 Daltons worth of input.
Your efficiency seems to be going the wrong way.

Also, bear in mind that your mass efficiency goes as the square root of your energy density. You need to quadruple the energy density in order to be able to cut the mass fuel rate in half.

Deleted member 690984
jbriggs444 said:
This new fuel is 22.4 Mev out of 8 Daltons worth of input.
The old fuel is 18.3 Mev out of 5 Daltons worth of input.
Your efficiency seems to be going the wrong way.

Also, bear in mind that your mass efficiency goes as the square root of your energy density. You need to quadruple the energy density in order to be able to cut the mass fuel rate in half.
Ah, of course. I think even with the triple alpha process in place, the fuel mass required is prohibitively massive.

How about antimatter as a fuel? Say, deuterium, and antideuterium.

paulthomas said:
Ah, of course. I think even with the triple alpha process in place, the fuel mass required is prohibitively massive.

How about antimatter as a fuel? Say, deuterium, and antideuterium.

Once we get to a "photon rocket", the relationship is that ##E=pc##. The energy requirement (E) is proportional to the momentum requirement (p) times the speed of light (c). [We're out of the Newtonian regime and firmly in the relativistic regime now].

Your energy density (##e##) is now ##c^2## (e=mc^2 and all that).

So if you want ##130 \times 10^{12}## Newtons then multiply by ##c## to get the energy requirement per second. ##130 \times 10^{12} \times 3 \times 10^8 = 1.5 \times 10^{25}## watts.

If you want the mass flow rate, divide by c instead of multiplying = 433 metric tons per second.

All assuming that I've not bungled horribly.

Deleted member 690984
Backing up a second - 1,020 zettawatts, what's that equivalent to in joules?

paulthomas said:
Backing up a second - 1,020 zettawatts, what's that equivalent to in joules?
Does not compute. A watt is a unit of power. A joule is a unit of energy.

Power is a rate at which energy is expended. One Watt is one Joule per second. A 100 watt light bulb (back when they were incandescent) uses up 100 joules every second.

So one zettawatt is one zettajoule every second.

Deleted member 690984
jbriggs444 said:
Does not compute. A watt is a unit of power. A joule is a unit of energy.

Power is a rate at which energy is expended. One Watt is one Joule per second. A 100 watt light bulb (back when they were incandescent) uses up 100 joules every second.

So one zettawatt is one zettajoule every second.
If I have it right, then the engines would need 1,020 zettajoules? Assuming that the antimatter annihilations aren't just fired out as thrust but power some kind of sci-fi engine, so isn't a "photon rocket" so to speak.

Am I correct in that a half-gram of matter and a half-gram of antimatter would produce 21.5 kilotons of energy? I read that 1 kiloton is about 4.184 terajoules, so a 1g annihilation would produce 89.956 terajoules, correct? So for that kind of energy, you'd need 11,300,000,000 grams, or 11,300 metric tons, correct?

paulthomas said:
If I have it right, then the engines would need 1,020 zettajoules?
Right. That's the 1.5 zettawats per engine times two engines times 300 seconds for a total of 1020 zettajoules.

paulthomas said:
Assuming that the antimatter annihilations aren't just fired out as thrust but power some kind of sci-fi engine, so isn't a "photon rocket" so to speak.
If it is not a photon rocket, what is it? Does it work by hurling out reaction mass? If so, it is a photon rocket.

If it works by grabbing hold of some sort of cosmic fabric and thrusting against it then that takes much less energy but it flies in the face of the principle of relativity. Something that physicists are unwilling to part with.
paulthomas said:
Am I correct in that a half-gram of matter and a half-gram of antimatter would produce 21.5 kilotons of energy? I read that 1 kiloton is about 4.184 terajoules, so a 1g annihilation would produce 89.956 terajoules, correct? So for that kind of energy, you'd need 11,300,000,000 grams, or 11,300 metric tons, correct?
E=mc^2. One gram is .001 kg. Times 3 x 10^8 squared is 90 terajoules. Yes.

But now you are mixing scenarios. If you have an anti-matter engine, why not throw photons out the back? Why carry inert reaction mass that will be thrown out the back when you could just haul some more antimatter and have yourself a photon rocket.

And if you are not using helium three fusion, why are you throwing reaction mass out the back with an exhaust velocity characteristic of helium three fusion?

Deleted member 690984
paulthomas said:
How about antimatter as a fuel? Say, deuterium, and antideuterium.
Are you intending that your ship is physically possible with current engineering and physics, @paulthomas? I'm assuming not, in which case you'll have to invent the parameters of your sub-light engines. In that regard, pretty much anything that requires reaction mass will be an engineering struggle, so it may be worthwhile considering reactionless engines and assigning them an acceleration value sufficient to get the ship moving to your top speed in a reasonable amount of time (which is likely hours rather than minutes or days). Or if you don't have artificial gravity, you can just set the acceleration to one gee and work out how long it takes to come up to speed.

Deleted member 690984
Melbourne Guy said:
Are you intending that your ship is physically possible with current engineering and physics, @paulthomas? I'm assuming not, in which case you'll have to invent the parameters of your sub-light engines. In that regard, pretty much anything that requires reaction mass will be an engineering struggle, so it may be worthwhile considering reactionless engines and assigning them an acceleration value sufficient to get the ship moving to your top speed in a reasonable amount of time (which is likely hours rather than minutes or days). Or if you don't have artificial gravity, you can just set the acceleration to one gee and work out how long it takes to come up to speed.
No, the ship isn't physically possible with current engineering and physics. I'm trying to make it as grounded as possible, but there are still some super-technologies in there, such as the ship's warp drive.

I think I may have to go with what you suggested, go with a reactionless engine of some kind, as using any kind of reaction mass is just going to slam head-first into the wall that is physical engineering difficulties.

Melbourne Guy

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