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:

Δwhere:v=uln [ (M+m) /M]

ΔNow in general, to get from one place to another a ship must accelerate for some timev= change in ship velocity

u= exhaust velocity

M= ship mass, without including reaction mass

m= reaction mass ejected from ship

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

Rearranging to give the required exhaust velocity:S= (T/2 + τ) Δv

= (T/2 + τ)uln [ (M+m) /M]

Now the poweru=S/ [ (T/2 + τ) ln [ (M+m) /M]

But the mass-loss rate is simply the total reaction massP= 1/2 ( dm/dt)u^{2}

This expression gives the maximum engine power required to propel a ship a total distanceP=m u^{2}/ 2T

=m S^{2}/ { 2T(T/2 + τ)^{2}[ ln [ (M+m) /M] ]^{2}}

d/dwhich has roots atT= 3/4T^{2}- 8t T+ 4t^{2}

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.T= 2/3t,

τ = 1/3t.

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 S^{2}/ [ 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).

Material | Volume/tonne (m^{3}) |
---|---|

Water | 1.00 |

Cadmium | 0.1156 |

Lead | 0.0881 |

Depleted Uranium | 0.0528 |