OK, we've all read the sci-fi tales of
great ships of the void zipping hither and yon at extreme speeds,
whether it's Doc Smith or David Weber.
Often, the ships must accelerate to
some large fraction of lightspeed using some form of conventional
propulsion, before being able to zap into hyperspace. Or, they just go Really Fast in normal space.
So... just what are the energy
requirements here? How much of the ship has to be fuel, and how fast
can it go?
For simplicity, I'll examine the
as-built Skylark, in which Richard Seaton went gallivanting off to
Osnome. (This is based on the heavily-revised 1958 Pyramid edition
of The Skylark of Space, 'cause that's the one I've got in
front of me.) This is a fairly simple one, for various reasons:
- Relativistic effects don't apply,
for reasons that are hand-waved away.
- The propulsion system uses total
conversion of matter to energy, which saves me the trouble of
looking up the energy yield of fusion reactions.
- The propulsion system is also,
presumably, perfectly efficient, and apparently reactionless.
- The mass of fuel is stated in the
book.
So, let's see now: “It was a
spherical shell of hardened steel of great thickness, some forty feet
in diameter.” Weight, “thousands of tons.” The energy source
is total conversion of copper: four-hundred-pound bars thereof; with
four bars aboard, the energy of one is available for outbound
acceleration, so we'll use that. When the Skylark is rebuilt
with an arenak hull, the “great thickness” of the original is
revealed to be four feet.
OK, now, we've got some numbers we can
work with. Converting to real units, we have a steel spherical shell
of outside radius 610 cm, and inside radius 488 cm, for a volume of
464 million cm3; taking the density of steel to be 7.86
g/cm3, that works out to 3.65E+6 kg, or 3650 (metric)
tons, for the hull, disregarding portholes and the massive internal
structures.
To accelerate this, we have the energy
a 400-pound copper bar, converted entirely to usable energy;
1.814E+2 kg converts, using the famous formula, to 1.63E+19 Joules.
So, when a 3.65E+6 kg object has a
kinetic energy of 1.63E+19 J, how fast is it going? Last I heard,
E=½mv2, so v=sqrt(2E/m) = sqrt(8.96E+12 J/kg) = 2.989E+6 m/s.
Just about 0.01c. Well, I guess we
didn't have to worry about those relativistic effects after all.
And so much for gallivanting around the
galaxy.
Note that none of this applies if the
hero has some sort of inertialess, or low-inertia, drive, such as a
Bergenholm; with those, the mass of the ship is eliminated or
reduced, so that kinetic energy ceases to be a consideration (or at
least is scaled down).
Update, a few minutes later: Corrected a couple of numbers; I forgot to apply the 2 in sqrt(2E/m) the first time around.
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