From The Exodus Project
Everything begins with the rocket equation:
Dv (km/s) = ve (km/s)*Ln[Mship (tons)/(Mship (tons) - Mp (tons))]
where:
Knowing Dv, we can approximate the acceleration produced by:
a (km/s^2) = Dv (km/s)/Dt (s)
where Dt is the time of the burn, in this case, the length of one turn. For reference, the acceleration due to gravity near the surface of the Earth is 1 g = 9.81 m/s^2 = 0.00981 km/s^2.
Knowing the acceleration allows us to find the distance travelled in one turn. This value sets the distance scale for the game (typically 1" on the board will equal this value). In real physics, the relationship is:
s (km) = 1/2 * a (km/s^2) *(Dt (s))^2
Note that here we've assumed a constant acceleration, which is clearly not the case, but the approximate formula for the acceleration used above is the average acceleration, which is (by definition) constant over one turn.
Now, our first departure from real physics: the movement systems of many games assume that the rocket reaches its final velocity instantaneously and then travels at that velocity for the entire turn. This is referred to as an impulse-type game. For these types of games, instead of using the average of the initial and final velocities (the initial velocity is assumed to be zero), we use the final velocity only. The effect of this is that the distance equation becomes:
s (km) = a (km/s^2) * (Dt (s))^2
In other words, we can travel twice as far in one turn if we assume we get all our velocity increase at the beginning of the turn rather than letting it build up over the course of one turn.
The only remaining question - and it is at the heart of rocket theory - is how much propellant are we burning each turn? The paradox of rocketry is that in order to get the maximum acceration possible from the rocket, we want to burn as much mass as possible and expell it at the highest exhaust velocity possible, yet we also want to conserve fuel! The two goals are of course mutually exclusive, so the game becomes to find the best balance between the two. We will not want the highest thrust engines (chemical rockets) because their fuel efficiency is too low to be of use for long-duration spaceflight. We also will not want the most fuel efficient engines (photon, or antimatter rockets) because the do not produce enough thrust to be usable in combat.
In general, one can specify the performance parameters of a rocket by providing the thrust (usually in kN) of the engine and the specific impulse (Isp, usually in ksec). The thrust is simply how hard the rocket pushes and is familiar to most space gamers. The Isp is, by definition, the impulse (force * time) of the engine produced by 1 kg of fuel. With the force (the thrust) measured in kN and the time measured in seconds, the astute reader will note that the units of Isp should be km/s (the same as velocity). In fact this definition gives the exhaust velocity of the rocket in km/s. For reasons lost in tradition, however, the Isp is reported in ksec, therefore it has been divided by the acceleration due to gravity (in m/s^2) so as to give the desired units. The actual definition of Isp is therefore:
Isp (ksec) = F (kN) * Dt (s)/(Mp (tons) * 1000 (kg/ton) * 9.81 (m/s^2))
and remember that:
ve (km/s) = Isp (ksec) * 9.81 (m/s^2)
Solving the Isp definition for the mass of propellant burned (Mp) gives:
Mp (tons) = F (kN) * Dt (s)/(Isp (ksec) * 1000 (kg/ton) * 9.81 (m/s^2))
or, equivalently:
Mp (tons) = F (kN) * Dt (s)/(ve (km/s) * 1000 (kg/ton))
We are now in a position to calculate the acceleration produced (and therefore the distance travelled) by any single rocket engine. The is no reason, however, why we can't have more than one engine aboard our ship. Adding extra engines does not change the exhaust velocity of the rocket, but it does increase the thrust produced - and therefore it increases the amount of propellant burned. If there are N identical engines on board the rocket, the mass of propellant burned is simply:
Mp (tons) = N (drives) * F (kN) * Dt (s)/(ve (km/s) * 1000 (kg/ton))
Now, how many engines can be placed on board a ship? This issue is decided by the design system (if any) in the rules of the game. For most systems, an engine (and its fuel) takes up a certain percentage of the total mass of the ship for each unit of acceleration ("thrust point") it produces. The mass of one engine as specified in the design system is the mass of the engine itself plus the fuel carried onboard. The next question is how much fuel is carried on board? We will assume that the ship must carry enough fuel to burn at maximum acceleration for the entire length of the game. How many turns, T, a game will continue will depend upon the game system in question. The total mass of one drive unit is:
Md (tons) = Mengine (tons) + T (turns/game) * Mp (tons/turn)
or
Md (tons) = Mengine (tons) + T (turns/game) * F (kN) * Dt (sec/turn)/(ve (km/s) * 1000 (kg/ton))
If the percentage of total ship mass allowed for engines (per thrust point) is P (e.g., 5%), the number of drive units we can have on board is:
N (drives) = P (percentage) / 100 * Mship (tons) / Md
or
N (drives) = P (percentage) / 100 * Mship (tons) / (Mengine (tons) + T (turns/game) * F (kN) * Dt (sec/turn)/(ve (km/s) * 1000 (kg/ton)))
The game system, then, must provide the values of P and T and either Dt (the length of one turn in secs) or s (the distance travelled in one turn in km) must be chosen. In general it is easier to specify the turn length and calculate the distance scale, but equations will be provided for the reverse. Notice our second departure from reality here: we have assumed that engines are infiinitely scalable, that is I can have "1.53 drive units". In actuality, some engines are simply to large to fit on a small spacecraft. If you want to be truly realistic, you should restrict yourself to using integer values of N. This means that under most design systems you will have search for the breakpoints in design which allow you to use integer numbers of drive units and smaller ships may have to use a different drive than larger ships.
We now have everything we need. The procedure is therefore:
From the game system, specify P and T. Note that few game systems have a hard limit on the number of turns in a game, so you must use your own experience to judge how long a typical game will last.
Set (arbitrarily) the turn length in seconds.
Find the mass in tons, the thrust in kN, and either the Isp in ksec or the exhaust velocity in km/s for your candidate engine.
Calculate the number of drive units onboard using:
N (drives) = P (percentage) / 100 * Mship (tons) / (Mengine (tons) + T (turns/game) * Dt (sec/turn) * F (kN) / (ve (km/s) * 1000 (kg/ton)))
Note that the mass of the ship is going to cancel, so it is really irrelevant unless you wish to install extra fuel tanks (see the fuel consumption rules elsewhere).
Calculate the mass of propellant burned in one turn using:
Mp (tons/turn) = N (drives) * F (kN) * Dt (sec/turn) / (ve (km/s) * 1000 (kg/ton))
Calculate the acceleration produced using:
a (km/s^2) = ve (km/s) / Dt (sec) *Ln[Mship (tons) / (Mship (tons) - Mp (tons))]
Record this number, it is the acceleration produced by 1 "thrust point". Divide by 0.00981 km/s^2 to get the acceleration in g's.
Calculate the distance travelled in one turn using:
s (km) = a (km/s^2) * (Dt (s))^2
If your game system allows you to use real physics, halve this number.
That's all there is to it!
The Full Thrust Fleet Book specifies that each thrust point (therefore all our results will be per thrust point) of the ship requires 5% of the total mass, therefore P=5. Typically, a Full Thrust game will not extend more than 10 turns, therefore T=10.
Many people have noted that it is desirable to have the Full Thrust and Dirtside II time scales match. Therefore, assume that one turn is 15 minutes, hence Dt = 900 sec.
Let's assume Full Thrust engines are deuterium-tritium fusion engines. The relevant parameters are: Md = 10 tons, ve = 22 km/s, F = 108 kN.
The number of drives which can be put in a 1 MASS (100 ton) ship (remember the mass is irrelevant) is:
N (drives) = 5 / 100 * 100 (tons) / (10 (tons) + 10 (turns/game) * 900 (sec/turn) * 108 (kN) / (22 (km/s) * 1000 (kg/ton)))
→ N = 0.09228 drives
Notice that this is (far!) less than one, so technically this engine will not fit on a 100-ton ship.
The mass of propellant burned in one turn is:
Mp (tons/turn) = 0.09228 (drives) * 108 (kN) * 900 (sec/turn) / (22 (km/s) * 1000 (kg/ton))
→ Mp = 0.4077 tons/turn
The acceleration produced is:
a (km/s^2) = 22 (km/s) / 900 (sec) *Ln[100 (tons) / (100 (tons) - 0.4077 (tons))]
→ a = 0.0000999 km/s^2 = 0.09987 m/s2 = 0.01 g
This is the acceleration produced by one thrust point.
The distance travelled in one turn using one thrust point is:
s (km) = 0.0000999 (km/s^2) * (900 (s))^2
→ s = 80.89 km/MU
Notice that if we had used real physics instead of the modified system used in the game, we'd get half this value.
I have written a Java applet to calculate the performance of several realistic engines or any engine you can come up with. The engines presented here are typical engines which we can either build now or feel we could in the relatively near future. In order to give everyone a common frame of reference, all of the values are taken from Phil Eklund's excellent Rocket Flight game. The few of these engines that I have been able to verify have been dead-on, so I'm willing to accept his values for the rest. If anyone has better information, I'd very much like to see it!
Full Thrust is a copyright of Ground Zero Games, no challenge to their copyright is intended.
Article ©1999 Keith Watt
All rights reserved.