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.
INTERPLANETARY TRIPS:
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...)
then
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 interstellar 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.
ORBITAL MECHANICS:
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 = ---------
r-squared
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 = -------------
r
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 = -------
r
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
631.457
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 (GM/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 8 times, T
increases 4 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 smaller than the
circular orbit distances:
Ring hexes/turn round off to (hexes/turn)
1 (6 hexes) 0.3618 1/3
2 (12 hexes) 0.2558 1/4
3 (18 hexes) 0.2089 1/5
4 (24 hexes) 0.1809 1 every 6 turns alternating
with 1 every 5 turns.
5 (30 hexes) 0.1618 1/6
6 (36 hexes) 0.1477 1/7
7 (42 hexes) 0.1367 1/7
8 (48 hexes) 0.1279 1/8
9 (54 hexes) 0.1206 1/8
10 (60 hexes) 0.1144 1/9
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.
ELLIPTICAL ORBITS:
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 [GM/P (1+e)],
and for apogee,
v(A) = square root [GM/A (1-e)].
You can see that these speeds are respectively faster and slower than
what is needed for circular orbits at those distances (GM/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.
