Sci-fi engine thrust calculations

[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?

Any help you can give would be greatly appreciated!

 

[Melbourne Guy]

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?

 

[PeroK]

Any help you can give would be greatly appreciated!

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.

 

[jbriggs444]

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.

 

[Deleted member 690984]

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.

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.

Reference: https://www.physicsforums.com/threads/sci-fi-engine-thrust-calculations.1017438/
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.

 

[Deleted member 690984]

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.

 

[PeroK]

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]

$5 \times 10^{9}\ \text{kg}$ 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.

 

[Deleted member 690984]

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.

 

[Deleted member 690984]

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.

 

[PeroK]

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.

There are the non-relativistic and relativistic rocket equations:

https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

https://en.wikipedia.org/wiki/Relativistic_rocket

 

[Deleted member 690984]

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!

 

[Deleted member 690984]

Yes, fiction is easier than mathematics!

Indeed! Are you able to help with my above problem at all?

 

[PeroK]

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."

 

[Deleted member 690984]

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?

 

[PeroK]

I want to make it as close to reality as I can. Can you help or not?

The reality is that we are earthbound.

 

[Deleted member 690984]

The reality is that we are earthbound.

So no, you can't, or do not want to. Please stop wasting my time.

 

[PeroK]

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!

 

[Deleted member 690984]

I know nothing about aneutronic fusion reactors, so no I can't help!

Then just say so to begin with, instead of being obtuse.

 

[jbriggs444]

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]

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)?

 

[jbriggs444]

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]

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?

 

[Deleted member 690984]

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?

 

[jbriggs444]

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.

 

[jbriggs444]

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]

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.

 

[jbriggs444]

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.

Your energy requirements will skyrocket. But your mass efficiency will improve.

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?