This is a second attempt for this question, the first time i put in too many different things that ended up derailing the question.

I'm in the early stages of designing a scifi game and want to use as little handwavium as possible (without ruining the fun). The first part of the game takes place in the asteroid belt and includes handwavium in the form of being able to break down just about anything into individual atoms and using a 3d printers make everything from an ingot to a fighter craft.

Backstory: Your ship was sent in advance to build the infrastructure for our first colony at Proxima b. The follow on ship with the colonists was scheduled to arrive 1 year after you but after 2 years the captain decided to head back and find out what was going on. FTL comms don't exist and the experimental FTL drives that we were used maxes out at 40c (just over 1 month to get to Proxima Centauri and only works in interstellar space so the crew was all in cryo sleep for the trip. When you wake up you find the ship is badly damaged and floating in the asteroid belt.

The intended reaction mass is sabots with a shell of (~11.75% mass) Invar (FeNi 64/36% 8,100kg/m³ specific heat capacity peaks at 545j/kg K at 200C Curie temp of 230C) wrapped around (~88.25% mass) silica harvested from asteroids. Depending on the ship the blocks may be as small as 0.01m³ ~30kg or as large as 1m³ ~3,000kg.

The smallest drone has a single 5m long coil, followed by dual 10/15/20m, quad 20m, and quad 50m for the largest coils. The dual and quad 20m fire 0.1m³ (300kg) slugs, the 50m fires 1m³ (3 000kg) slugs, everything else is 0.01m³ (30kg). The power budget is 56MWh for the smallest, 112MWh for the duals, and 278MWh for the quads. All coils are up to 5m outer diameter. Oh, and these are only used in space so air friction isn't an issue.

So, finally, the question is how do i turn all that into newtons for thrust/deltaV calculations?

Edit: I found Luke Campbell's coilgun equation on several different coilgun questions. http://www.projectrho.com/public_html/rocket/spacegunconvent.php#id--Kinetic_Kill_Weapons--Coil_Guns If i did the math right 398098J is enough to push a 3000kg slug to 1000m/s in a hair over 59 meters. Did i get it right?

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    $\begingroup$ Invar's a very, very strange choice for reaction mass pellets. It's defining trait is a low coefficient of thermal expansion, which is pretty much irrelevant here. The properties you're interested in are electrical conductivity (about 5 thousand times worse than copper) or its magnetic properties (unimpressive permeability, fairly low Curie temperature, etc). And do you have a particular reason for manufacturing precision projectiles for reaction mass instead of just slinging mass out of a bucket which is braked within the mass driver and reloaded for another shot? $\endgroup$ Sep 10, 2022 at 0:28
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    $\begingroup$ Your mass drivers are also rather short, and the projectile energies are quite out of scale with the projectile masses...your smallest drivers are nearly 5 times as efficient in their use of reaction mass, which raises the question of why the larger ones even exist. Anyway, the average thrust is determined by the average mass flow rate and the exit velocity of the mass driver. Projectile composition, density, temperature, mass, size...none of these matter. $\endgroup$ Sep 10, 2022 at 0:40
  • $\begingroup$ Ah, but they do matter because some of the energy you put into the magnetic field will become waste heat. The higher the temp you can heat the slug the more energy you can impart. As for why invar, the is in the asteroid belt where most iron you find will have quite a bit of nickle already in it unlike deposits on Earth. $\endgroup$
    – Rasip
    Sep 10, 2022 at 0:58
  • $\begingroup$ The different sized mass drivers and slugs exist because they are pushing vastly different sized ships. Everything from a 8x16m drone to a 50x105m main module to multiple 50x70m modules. While yes there are vastly better options available these are made from the most common materials found in the asteroid belt. $\endgroup$
    – Rasip
    Sep 10, 2022 at 1:05
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    $\begingroup$ I suspect that if you really optimize for the most efficient projectile to launch from an electromagnetic engine, you'll just end up recreating ion engines. $\endgroup$
    – Cadence
    Sep 10, 2022 at 2:00

3 Answers 3


Newton's third law can be written (using the 2nd law) as $\vec F_1 = m_1 \times \vec a_1 = -\vec F_2 = m_2 \times a_2 $ -- use that and conservation of momentum.

Each block is thrown at some velocity $\vec v$, so has a momentum of $\vec p = m \times \vec{v}$ -- that means it gives the same momentum (in [kg m/s]) to the ship that tossed it. But momentum $\vec p$ is also $\vec p = \int \vec F dt = \vec F \times t$, so you calculate the length of time it takes to accelerate the block (assume half final velocity over the length of the driver -- it'll be close enough) and back out the force required to get that velocity in that time.

Reaction on the drive (hence the ship) will be the same magnitude in the opposite direction for each single block. Then calculate how many blocks per second (or other convenient time interval) you need to toss to get the overall acceleration you want/need as well as how long your reaction mass (i.e. store of premade blocks) will last while accelerating.

Knowing the force (either for a single shot or averaged over time) will get you the acceleration of the ship after dividing by the mass of ship plus remaining reaction mass store. The rocket equation will give you the total delta-V you can get from a given capacity of blocks or the actual delta-V obtained by using any fraction of the available reaction mass.

  • $\begingroup$ you used 2nd law formula, fixed it to the composite 2nd/4rd law variant. 2nd is "F=ma" and 3rd law is "action = reaction" $\endgroup$
    – Trish
    Sep 14, 2022 at 14:26
  • $\begingroup$ Thanks, @Trish. I never bothered to write those as vectors for calculation purposes, but that's surely more correct than treating them as scalars -- though I'm not sure how relevant it is to this question. $\endgroup$
    – Zeiss Ikon
    Sep 14, 2022 at 14:33
  • $\begingroup$ it's basically irrelevant because the question is along one axis, but it might be relevant in case the railgun (or in case of multiple: their averaged out centerpoint) is not through the center of mass of the ship - that'd create a turning momentum $\vec M$ $\endgroup$
    – Trish
    Sep 14, 2022 at 14:41
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    $\begingroup$ And unless that's due to damage -- say from a minor collision -- you'd then have to go all the way back to Earth to shoot the engineers... $\endgroup$
    – Zeiss Ikon
    Sep 14, 2022 at 15:06
  • $\begingroup$ ...or one engine failing, at which point engineers are to be shot too! $\endgroup$
    – Trish
    Sep 14, 2022 at 15:09

let me cut to the needed parts...

The smallest drone has a single 5m long coil, followed by dual 10/15/20m, quad 20m, and quad 50m for the largest coils. The dual and quad 20m fire 0.1m³ (300kg) slugs, the 50m fires 1m³ (3 000kg) slugs, everything else is 0.01m³ (30kg). The power budget is 56MWh for the smallest, 112MWh for the duals, and 278MWh for the quads. All coils are up to 5m outer diameter. Oh, and these are only used in space so air friction isn't an issue.

The composition doesn't matter, the material doesn't matter. All that matters is the mass, weight and ejection force.

  1. $\vec F=m\vec a$ - Newton's 2nd law
  2. $\vec F_1=-\vec F_2$ - Newton's 3rd law "action = reaction"

F is measured in Newtons.

You give us the devices' power budgets in MWh, which is actually easily convertible into their energy in Joules: 1 kWh is 3.6 MJ, so 1 MWh is 3600 MJ, your smallest drive brings 201 600 MJ to the table. That is the Work the device delivers. Assuming 100% efficiency, we can use that as the work that is used to accelerate the projectile. Work is defined as Force times the distance it is moved:

  1. $|\vec F|=F$ & $|\vec a|=a$ - For simplicity, we use the length of the force and acceleration, as we look at a single axis: the one along the coilgun. If that is not through the center of mass, you need to calculate a lot more... later.
  2. $W=F s$ - Work (is a scalar, not a vector!)

So, we have everything we need to solve the force each shot does have!

  1. $F=W / s=\pu{201600 MJ} / \pu{5m}=40320\times \pu{10^3 \frac{kg\ m}{s^2}}$
  2. $a_0=F / m=40320\times \pu{10^3 \frac{kg\ m}{s^2}} / \pu{300 kg} = 134.4 \times \pu{10^3 \frac{m}{s^2}}$

$1.3\times \pu{10^5 \frac{m}{s^2}}$ in lower than the acceleration you get from a conventional gun. So, we have the (linear) acceleration the projectile gets. From that we can calculate the time it has in the barrel:

  1. $s=s_0+v_0t+a_0t^2$ - the basic movement formula, solve for time with known $s_0=0$ and $v_0=0$
  2. $t_{exit}=\sqrt {s_{exit} / a_0}=\sqrt {\pu{5m} / {134.4 \times \pu{10^3 \frac{m}{s^2}}}}=\pu{0.006099 s}$

So, our projectile has one 164th of a second in the device. How much momentum does it give the ship? Let's do some conversion math starting with

  1. $p=m\times v(t_{exit})$ - the standard formula for impulse, now we apply all those formulas 5 to 8 in reverse order:
  2. $p=m\times (a_0 t)=m\times (a_0 \sqrt{s/a_0})=m\times ( \sqrt{s \times a_0})=m\times \sqrt{s\ F/m}=m\times \sqrt{s\ (W/s)/m}=m\times \sqrt{W/m}=\sqrt{W \times m}$
  3. $p=\sqrt{\pu{201 600 MJ} \times \pu{30 kg}}=\sqrt{6048\times \pu{10^9 \frac kg^2m^2/s^2}}\approx\pu{\pu{77769 kNs}}$

Now, to see how fast your ship goes after each shot is simple!

  1. $m_{ship}(t)=m_{dry}+N(t)\times m_{projectile}$ where N(t) is the number of shots left at the moment t.
  2. $\Delta v_{ship}(t)=m_{ship}(t+1) / p$ - This is the speed change if you shoot one projectile at the time t.
  • $\begingroup$ Might just mention that the mass block is accelerating at about the same rate as a rifle bullet would in the barrel. 13000+ G. Hope that Invar doesn't soften much as it heats, or you might have a problem with the buckets stripping off the stone. $\endgroup$
    – Zeiss Ikon
    Sep 14, 2022 at 16:14
  • $\begingroup$ @ZeissIkon yeeea, no. 134 kilometers per seconds-squared over 0.006 seconds... we are accelerating less harsh than a normal projectile. $\endgroup$
    – Trish
    Sep 14, 2022 at 16:26
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    $\begingroup$ @Rasip "Those power budgets are the entire amount the batteries store (preferably using less than half to accelerate and the other half to decelerate), not the budget for each shot.": that's something of a problem: your largest 1 TJ battery contains a pitifully small amount of energy when it comes to moving a spacecraft. $\endgroup$ Sep 14, 2022 at 21:10
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    $\begingroup$ @Rasip Basically, if by "battery" you mean an electrochemical device producing electricity from chemical reactions, this just isn't going to work. Such a thing just isn't going to approach the energy density of tanks of chemical propellant even on its own, it'll be far worse when you account for the reaction mass being separate. Batteries could be useful for accumulating the energy needed for each shot from some lower-power energy source like a reactor or solar array, but not for storing the energy needed for an entire interplanetary flight. $\endgroup$ Sep 16, 2022 at 14:04
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    $\begingroup$ @Rasip I think you're going to want to drastically rethink how you've scaled things. A 1 TJ NiMH battery will mass over 2 thousand metric tons. Launch smaller projectiles more often and you can reduce the battery size. $\endgroup$ Sep 17, 2022 at 5:28

More as a comment:

  • handwavium in the form of being able to break down just about anything into individual atoms and using a 3d printers make everything from an ingot to a fighter craft.

Not so much of a hanwavium, I would say. All 3d printing is a bit a handwavium, but also it depends on tech, and the angle we define it as all 3d printing - as 3d printing tools of production to start actual production also can be considered 3d printing in some way.

As for the q, Trish way too complex much(but kudos for efforts, that for sure).

Things are simple, what is conserved is an impulse, meaning mass multiplied by change of velocity, which, the change of the velocity, at its max is known as delta-v.

Conservation means that mass of a ship multiplied by ship velocity change is equal to projectile mass multiplied by its exhaust velocity.

And as rough estimation all there is needs to know is exchaust velocity of a projectile, which does not depends that much on reactor power output, but a function of design of the whole system (capacitors and such can reduce demand on max reactor power output and it will be, most likely, an improvement as it may save mass thus imroving delta-v and acceleration time)

If reactors are soo good at delivering 56MW for small drones, then it makes sense to use ion based engines, to which any asteroid mass can be a reactive mass. You can use everything you do not need in this asteroid(oxygen as an example, which about 40-60% of asteroid mass) to be expelled for purposes of reactive propulsion.

And in a sense ION engines are mass drivers, but with a higher exchaust velocity of reactive mass and thus higher delta-v.

It may be ok to have not a high delta-v engines, being in asteroid belt - depends on curcumstances and constraints purposes.

As side note, 1km/s slugs are quite low velocity, which means also low delta-v which may be fine in a belt, to get from one worked out asteroid to another, but it will take its time, as they quite sparse placed(say no to starwars asteroid belts!) and in that sense ION engines are not loosing anything compared to those railgus - they save more mass for you and won't have worse time for moving from one asterid orbit to another, so as they are fine for hopping from close asteroids chunks 1-10km apart (sure it slower ther but, it not how you establish a connection between those two chunks on regular basis, so it is fine for some particular cases)

As means to deliver asteroid mass to some main asteruid factory, those railguns, mounted on an asteroid, can be viable option to gather mass to be processed. But it needs to evaluate what can be done from one cubic km asteroid, 1% iron content from it means about 0.5% of this planet annual steel production, industry for few billon people, so that 1% of asteroid stuff is something on the level of an industrial city of 10's of million people. So one asteroid of that size can be quite self sufficient productiin facility which does not need any external materials delivered, or moved away.


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