Reaction Drive Formulae

The rules given in GURPS Space for reaction drives are okay as far as they go, but they do not give sufficient detail for GMs wishing to make reaction drives a major part of a campaign background. The emphasis in GURPS Space is on reactionless thrusters, which make starship design calculations quite simple. For a sub-light drive which doesn't violate the laws of physics as we know them, however, we must use reaction drives.

The power and reaction mass requirements of reaction drives are governed by non-trivial relationships. This page derives the relevant formulae and discusses their consequences for a GURPS Space campaign. Warning: Some knowledge of algebra and basic physics is required to follow the derivations, but anyone should be able to use the end results and discussions.

First, two important definitions:
Fuel: The stuff that is used to generate power for the ship's drive. Not necessarily the same stuff that is ejected out the back of the ship. Examples: nuclear fuel rods, deuterium, antimatter.
Reaction mass: The stuff that is ejected out the back of the ship to provide propulsive force. Not necessarily the same stuff that is used to generate power. Examples: liquified hydrogen, water, cadmium.

We begin with the classical (ie. non-relativistic) rocket equation. For a derivation of this equation, see here. We use the classical version because relativistic effects only become important for exhaust velocities greater than about 95% the speed of light, which is not the case for the powers and speeds we will be discussing. The rocket equation is:

Δ v = u ln [ ( M + m ) / M ]

where:

Δ v = change in ship velocity
u = exhaust velocity
M = ship mass, without including reaction mass
m = reaction mass ejected from ship

Now in general, to get from one place to another a ship must accelerate for some time T /2, then coast at top velocity for a time τ, then decelerate over a time T /2. The total change in velocity is Δv, but since the ship speeds up and slows back down to rest, the maximum velocity is Δv /2. The total trip time t = T + τ, the time spent accelerating/decelerating plus the time spent coasting.

The total distance covered S is the area under this graph:

S = (T /2 + τ) Δ v
= (T /2 + τ) u ln [ ( M + m ) / M ]

Rearranging to give the required exhaust velocity:

u = S / [ (T /2 + τ) ln [ ( M + m ) / M ]

Now the power P required to eject the reaction mass at the given exhaust velocity is equal to the rate of change of kinetic energy of the reaction mass, which is half the mass-loss rate dm /dt times the square of the exhaust velocity:

P = 1/2 ( dm /dt ) u²

But the mass-loss rate is simply the total reaction mass m divided by the total acceleration/deceleration time T. Substituting:

P = m u² / 2T
= m S² / { 2T (T /2 + τ)² [ ln [ ( M + m ) / M ] ]² }

This expression gives the maximum engine power required to propel a ship a total distance S, and slow it down again, given known values for the ship mass, reaction mass, and desired acceleration/deceleration and coasting times. The engine uses no power while coasting, of course. For a desired total trip time t, we can minimise the maximum required power by maximising the denominator expression T (T /2 + τ)² subject to the constraint t = T + τ. To do this we substitute for τ and differentiate with respect to T, keeping t constant:

d/dT = 3/4 T² - 8t T + 4t²

which has roots at T = 2/3 t and 2t. The second solution is unphysical, but the first provides the mimimal engine power solution at:

T = 2/3 t,
τ = 1/3 t.

So to get somewhere in a given time, using a given reaction mass, but using the least possible engine power, accelerate for 1/3 of the time, coast for 1/3, and decelerate for the final 1/3 of the time.

For application of these calculations to ship drive and power system design, see the discussion of space drives.

Okay, so we've figured out how big our ship power plants need to be. This is campaign information, and should be designed by the GM before the campaign begins. During game play, however, the PCs will know how powerful their engines are, how far they are going, how much their ship masses, and how much reaction mass they have. The most important question is going to be: How long does it take us to get from here to there?

To solve this, we need to rewrite our last equation to make total trip time the subject. We have a free variable - the coasting time - but similar calculus to that shown above proves that the total trip time is minimised by accelerating for the first third of the trip, coasting for the next third, and decelerating for the final third. (Note this is only the case assuming we want to use a fixed amount of reaction mass. If we had reaction mass to spare, of course we could get there faster by accelerating halfway, then decelerating without a coasting period.) Under these conditions, the total trip time t becomes (this equation is shown in both text and graphics formats):

t = ( 3/2 ) { m S² / [ 2P ( ln [ ( M + m ) / M ] )² ] }^{( 1/3 )}

The following JavaScript form calculates this minimum trip time for you. The ship mass does not include the reaction mass. Be careful to enter your data in the correct units (as indicated).

Ship Mass (metric tonnes):

Reaction Mass (metric tonnes):

Engine Power (megawatts):

Distance (astronomical units):

This trip will take:

Reaction Mass Material

GURPS Space gives only three options for reaction mass material, and seems to have the density of cadmium wrong. Here are some other options, with their real volume/mass ratios.

Material Volume/tonne (m³)

Water 1.00

Cadmium 0.1156

Lead 0.0881

Depleted Uranium 0.0528

Material	Volume/tonne (m³)
Water	1.00
Cadmium	0.1156
Lead	0.0881
Depleted Uranium	0.0528

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