By Pierre Savoie [published with permission]
(People are free to edit and include these facts in any Web-page or CD-ROM text they want, however they want. )
The game scale for Star Frontiers' Knight Hawks (SF's KH) game consists of 10,000-kilometer (km) hexagons (hexes), and a game turn for starship movement is 10 minutes (600 seconds or 600 s). Now, the speed of light, c, is 299,792.5 km/s, and 1% of c works out to 179.8755 hexes/turn (10,792,530 km/h). Since any hex-based wargame is very "chunky", that is, ships are considered to be either at the center of one hex or at the center of another, naturally this can be approximated as 180 hexes.
It would be very hard to show this kind of speed on a board of finite tabletop size, although you could shift the frame of reference to the pursuing ships and show relative position, i.e. one of the pursuing ships stays still on the map and the other ships move relative to it, noting real speeds of each ship, but only the DIFFERENCE in speeds counts to set a new relative position.
HOW LONG UNTIL WE CAN JUMP?
In Star Frontiers, safe jumps into hyperspace are not a matter of the ship being sufficiently away from the gravitational field of planets and stars. In SF any ship can jump once it reaches 1% of the speed of light, thus being able to access the Void with a Void Jump Engine. To find out how many turns this takes, divide 180 by the ADF (acceleration/deceleration factor, in units of hexes/turn/turn acceleration) which the ship uses to accelerate out. For ADF 1 this takes 180 turns or 30 hours. For ADF 5 this takes 36 turns or 6 hours. However, the time for calculations must also be considered if greater than the time physically necessary to accelerate. The astute person might ask, "relative to what?" Planets, stars and even galaxies move, and all this adds up to giving the ship an inherent movement of a few dozen kilometers per second relative to some extra-galactic frame of reference even when it is standing still on a planet. This is negligible compared to 1% of c, however. In addition, you could rule that the speed achieved should be relative to the local source of gravity, whether a sun or a planet if near one.
KH GAME MOVEMENT:
The rules for the KH game have two basic flaws in them. The MF or maneuverability factor rule makes possible 180-degree turns, apparently regardless of ship's speed. It would take many more turns to slow and stop a ship and speed it up in the opposite direction by the same amount, yet the equivalent action of turning 180 degrees can be done in the same turn if a ship is sufficiently "maneuverable". Physically, this is impossible, since the amount of thrust needed to alter a ship's course in both cases must add up to the same. The only way this is possible is if ships are equipped with some sort of "blue sky" (fanciful, futuristic) device that plays with the nature of inertia itself, but affecting direction more than speed (?).
The second problem is that, for boardgame simplicity, ships are assumed to be able to change their speed at the start of a turn, and travel a number of hexes in that turn equal to their new speed. But on the present scale this is difficult. Even 1 ADF as an acceleration spread out over the whole 10-minute turn represents a speed change of (1 hex/turn)/turn, or (10,000 km/600s)/600s, or an acceleration of 27.777777... meters/second (m/s). Since acceleration of a falling object due to gravity on Earth at sea level and 45 degrees latitude is 9.80665 meters/second/second (m/s/s), 1 ADF applied throughout the turn represents 2.8325 gravities or G. This is increased slightly as you go to the poles and decreased as you go to the equator, from the varying distance from the Earth's center. In addition to this, the apparent "weight" of a non-falling object varies slightly with the centrifugal force of the Earth's rotation, which varies from zero at the poles to a maximum at the equator.
But 2.8325G (ADF 1) is very uncomfortable to humans with no special equipment, and ADF 5 is extreme, probably fatal over extended periods. Also, after 10 minutes the speed is a new speed in hexes/turn, but the distance travelled that turn under constant acceleration is the average of the initial speed and the final speed. The game merely assumes the new speed holds. A fighter accelerating from 0 to 5 hexes/turn in a turn has only in fact travelled 2.5 hexes, but the game rules say 5 (this is only true if the change in speed has all been bunched up in the very beginning of the turn). For added accuracy, the rule could be changed in which movement per turn is the average of the speed in hexes/turn at the beginning of the turn, and the speed at the end of the turn; this would cause little difficulty but would introduce half-hexes or cause a rescaling of hexes.
In the primitive past of Star Frontiers, spaceships were limited in terms of fuel and thrust that they could deliver. The reaction mass was the exhaust products of the fuel itself, heated by the reaction of the fuel to reach as fast a speed as possible that could be safely channelled out in a specific direction (without abrading the nozzle surfaces). Spaceships were basically like controlled, slowly exploding bombs. Often, spaceships were designed as stage-rockets, consisting of ejectable fuel sections to take advantage of improved thrust-to-weight ratios as empty, useless sections of the rocket were expelled, so that not all the ship had to be lifted completely out of a planet's gravitational pull. As reaction mass is expelled, the whole ship gets lighter yet the thrust is constant, so ships are accelerated faster the less fuel is left. This is easy to calculate using the ROCKET EQUATION, using a little calculus about the imparting of momentum to the ship by the hot gases thrust backwards from it:
If V(t) = speed of ship at time t, V(e) = speed of exhaust gases, M = total mass of ship at the beginning (of launch, or of a stage), M(t) = total mass of ship at time t, and ln represents the "natural logarithm" or logarithm to the base e (2.718281828459045...)
V(t) = V(e) * ln [M / M(t)] (+ any initial speed from a catapult or an earlier stage)
In the early days, fuel was the limiting factor in choice of orbits, as ships struggled to escape off the planet and achieve a stable orbit, then inject themselves into a parabolic or hyperbolic transfer orbit, to be captured by a target planet or moon, with an elegant minimum of fuel. Such an approach took time: an economical fuel burst at the beginning, a long wait, and then a nearly equal fuel burst to brake at the end. Total braking was not used, because a remaining speed was retained to enter orbit at a certain altitude at the appropriate speed (see below).
When faster spaceships were developed, economical orbits were less of a concern, and orbits were more direct, less elliptical and straighter. If reaction mass could be ejected at faster and faster speeds, less of it would be needed and so the mass of this reaction mass compared to the total mass of the ship got smaller for the same power. Ships could burn fuel (or heat and eject reaction mass) longer, often constantly, and so an interplanetary trip consists of accelerating until halfway and then decelerating the rest of the way.
Under these circumstances, if any two out of three variables such as travel time T, distance D and constant acceleration-then-deceleration A are known, the third can be calculated by the following formulas:
T = 2 * square root[D/A]
D = A * T-squared /4
A = 4 * D / T-squared
These formulae work for units of meters, seconds and meters/second/second. They ALSO work if the units are hexes, turns and ADF used, since the acceleration unit is directly derived from the scale units of time and distance. In practice, two factors cause slight differences to the time of travel needed:
a) normally, the ship already has some initial speed from orbiting a planet, so accelerating to that small amount is not required.
b) movement occurs in an environment of a varying gravitational potential well. Ships are slowed as they move away from the sun, and speed up if they approach a planet nearer the sun. Some correction for this is needed, and so generally more time is taken.
In the force of gravity, all objects attract all other objects with a force proportional to their mass and inversely proportional to the square of the distance between them. For small objects this force is negligibly small, but for large objects the size of planets, the force becomes significant. The universal formula for gravitation is:
k * M * m
F = ---------
where F is the force of gravitation, k is a constant, r is the distance between the center of a planet and the object, M is the mass of a planet, and m is the mass of the object the planet acts on.
In metric units of meters (m) and kilograms (kg), the gravitational constant k is 6.6720 x 10E-11 newtons m-squared/kg-squared (where E signals an exponent or "to the power of"). (A newton (N) is a unit of force which if applied is able to accelerate a 1-kg object with an acceleration of 1 m/s/s.) Since the Earth masses 5.9763 x 10E24 kilograms, and since acceleration due to the force of gravity, a(g) (in m/s/s), is the same regardless of the mass m of an object, this can be restated for the Earth as:
3.9874 x 10E14
a(g) = -----------------------
(r in meters)-squared
For distance r in kilometers, this is
3.9874 x 10E8
a(g) = -----------------------
(r in km)-squared
The Earth is 6,378.245 km in radius at the equator (and 21.5 km less at the poles), so this acceleration due to gravity is 9.801 m/s/s.
For a planet of different mass, the acceleration due to gravity varies proportionally to the new mass and in inverse-square proportion to the distance from the surface to the center. Thus, for planets with density d times that of the Earth, and a radius x times the radius of the Earth, they will have exactly xd times the surface gravity. Gas giants have about 1/5th the overall density of the Earth, but many times the radius. Jupiter, for example, is 317.8 times the mass of the Earth and 11.19 times its radius, but because of the lowered density its "surface" gravity is 2.69 times Earth's. The largest gas giants are 3 times the mass of Jupiter, because beyond that mass the extra gravity due to their weight compacts them back down to a smaller radius. If a gas giant is 70 to 80 times the mass of Jupiter, it may ignite into a small red dwarf star.
At the Earth's equator, if an object is 5 m above the ground, it would take just over one second to fall to the ground. If, however, it has a sufficient forward velocity, it will move forward and away from the Earth's curvature faster than it is falling towards it. Of course, it could not easily orbit within the Earth's atmosphere because the atmosphere will slow it down by friction. For that reason the lowest practical parking orbit is about 200 km above the surface to get away from the atmosphere.
When an object is in a circular orbit around a planet, a force of gravity is acting on it, dragging it down. But any object moving in a circular fashion must have a centripetal force acting on it, which from geometrical considerations equals
m * v-squared
F = -------------
where v is speed of the object in a circle in meters/second, r is the radius of that circle in meters, m is the mass of the object in kilograms.
But if we equate this force to the force in the universal gravitational equation from earlier, then
k * M
v-squared = -------
For the Earth's case, this can be restated so that the speed v in km/s needed to achieve a circular orbit at a given altitude a in km is
v = -----------------------
square root(6378 + a)
Thus for an orbit at 200 km altitude, the speed needed is 7.785 km/s.
The period of revolution T of an object in orbit is simply the time it takes to trace a circle at its circular orbital speed:
d 2 * pi * r
T = --- = -------------------------
v square root (k * M / r)
For the Earth's case, T in hours for an altitude a in km is:
T = 2.76397 x 10E-6 * (6378+a)E1.5 (E1.5 means "to the power 1.5")
Thus, for a geosynchronous orbit of T = 23.933 hours, satellites must be at an altitude of 35,789 km above the surface. A similar calculation can be made for Star Frontiers planets, if one assigns a suitable mass and radius for each planet.
In general, for orbits around the same planet or the same sun, T-squared is always proportional to r-cubed. Thus, if r is increased 4 times, T increases 8 times (i.e. T-squared and r-cubed both increased 64 times).
What does this mean for the scale of Knight Hawks? A planet of a size like the Earth occupies slightly more than one hex in width. A ship at the center of an adjoining hex is 10,000 km from the center of the Earth and orbits with a period of 2.764 hours. But the approximate KH distance it covers in this orbit (10,000 times 2 times pi or 62,831.8 km) is represented as 6 hexes, in 16.58 turns, so this has to round off roughly to 1 hex moved every 3 turns -- 3 times slower than the tactical KH rules suggest!
Here is a table showing orbital speeds in hexes/turn for each hex "ring" of 10,000 km from the Earth's center, and the KH game approximation that can be used. Some distortion occurs since the hex distances are crossed in straight lines, which are smaller than the circular orbit distances:
turn round off to (hexes/turn)
|1 (6 hexes)||
|2 (12 hexes)||
|3 (18 hexes)||
|4 (24 hexes)||
every 6 turns alternating with 1 every 5 turns.
|5 (30 hexes)||
|6 (36 hexes)||
|7 (42 hexes)||
|8 (48 hexes)||
|9 (54 hexes)||
|10 (60 hexes)||
For other planets, the orbital speeds for the same distances will vary in proportion to the square root of the variation in planetary mass. Thus, a planet four times as massive as the Earth requires doubled orbital speeds. The mass itself varies in proportion to the density and in proportion to the cube of its radius. So, keeping density the same, if a planet has four times the radius it has 64 times the mass, and all orbital speeds at a given distance from the center increase 8 times.
In practice, orbits are rarely exactly circular. Sometimes a transfer orbit from a low insertion-orbit to a higher orbit (such as for a geosynchronous satellite or a space-station) must be made. Such orbits are elliptical, and the center of the planet is one focus of this ellipse. These orbits vary between a perigee P (closest distance to the planet's center) and an apogee A (furthest distance to the planet's center). The "semi-major axis" of the ellipse, a, is the distance from the center of the ellipse to one of its longer ends, and a happens to equal (A+P)/2.
The ellipse's "eccentricity" e is equal to the distance between the center of the ellipse and the center of the planet, divided by the ellipse's semimajor axis. This turns out to equal
A - P
e = -----
A + P
The velocity at any point along the chosen orbit can be found by the equation:
v = square root [ k * M * (2/r - 1/a)]
where r is the distance from the center of the planet and a is the semimajor axis of the ellipse, which equals (A+P)/2.
For perigee, this works out to
v(P) = square root [kM/P (1+e)],
and for apogee,
v(A) = square root [kM/A (1-e)].
You can see that these speeds are respectively faster and slower than what is needed for circular orbits at those distances (kM/r). This is expected since an elliptical orbit at apogee does not quite have the speed to make a circular orbit, and so drops closer towards the planet or sun and thus gains speed. At perigee it has a faster speed than the circular orbital velocity needed, and so moves away from the planet in an elliptical path, slowing down until it reaches the apogee again.
At any point, if the velocity is suddenly changed to the original velocity times the square root of 2, the spaceship no longer travels in a closed path but reaches ESCAPE VELOCITY for that altitude. The eccentricity of the ellipse jumps to 1 and the path becomes a parabola. Slightly more than this speed will create a hyperbolic path which effectively escapes the planet and creates a new elliptical orbit around the sun, presumably to intercept a given planet and brake to assume an orbit around that planet.
I hope you have found this astrophysics discussion useful.
Pierre Savoie (email@example.com)