From: travis@cco.caltech.edu (Leopold E. Travis)
Subject: Re: Star Wars vs. Star Trek, tech & tactics, etc.
Organization: California Institute of Technology, Pasadena
Date: Tue, 28 Jul 1992 16:51:18 GMT


rlbell@babbage.waterloo.edu (Richard Bell) writes:
>travis@cco.caltech.edu (Leopold E. Travis) writes:
>>[Much of my speculation on the Feds vs. the Death Star deleted.]
>
>The first weapon strike near the gun will cripple it.  The slightest drop
>in energy efficiency will instantly destroy the Death Star the next time it
>fires.  The Death Star will be destroyed even sooner if the Feds can warp 
>space into a set of gravity lenses that send the beam back to its source.
>That might require some quantum black holes, but we can plainly see that
>the Feds have artificial gravity fields to help them manipute things.
>
>The feds will lose a few planets before they get it right, but the planets
>will be only outlying colonies (they will be found first in a search).
>Building a planet based Death Star busting phaser will require less effort
>than building a planet busting Death Star based weapons (Death Stars being
>such puny things in astronomical terms), so Fed planets will stop being 
>easy pickings.  If the Feds start giving them to the rebels, the Empire
>evaporates almost as quickly as the imperial fleet.

I'll address the issue of quantum black holes later in this post. 
First, let's examine the conflict assuming no new weapons technologies,
i.e. both sides are restricted to their standard armaments in the TV
shows and movies.  

Let's look at the vulnerability of the Death Star's gun.  As I've stated
before, it must have required about 10^17 megatons to destroy Alderaan. 
If this feat was accomplished in 10 seconds, this puts the Death Star's
energy output at 10^16 megatons per second (about 4 x 10^31 Watts, or
100,000 times the output of the sun (!)).  So if the mechanism of energy
release is matter-antimatter annihilation, the Death Star must consume
about 2.3 x 10^14 kilograms of antimatter (and an equal quantity of
matter, of course) per second when firing its gun.  Clearly this must
require an incredibly powerful containment field.  The machinery
involved must take up a good fraction of the Death Star's bulk, and
mostly be situated deep within it.  So surface photon torpedo hits are
not going to take it out.  Since the Death Star's radius is 60
kilometers and a photon torpedo only digs about 0.3 kilometers into it
(this was a conclusion of my previous post, assuming a yield of 100
megatons for a photon torpedo), it's going to take dozens of hits upon
the same spot to get at the bulk of the gun machinery. 

Furthermore, suppose the Feds do manage to damage the gun machinery. 
Surely this must be the most heavily backed-up and fail-safed system on
the Death Star, so it'll take a great deal of damage to cause it to fail
completely.  And even if the Death Star can only crank out a millionth
of its usual output, this is still 100,000,000,000 megatons per
second--easily enough to destroy all life on the target planet (and
vaporize the oceans and probably the crust as well; recall that the
arsenals of the United States and the Soviet Union during the cold war
were on the order of a "mere" 10,000 megatons each, and this would
probably have been sufficient to wipe out civilization as we know it).

Finally, suppose the Feds do take out the containment field, causing the
Death Star to explode.  Assuming it carries enough antimatter for 100
seconds of output from its gun (enough to destroy ten planets), this
means the blast is going to be about 10^18 megatons--which will still
wipe out the target planet and the attacking Federation ships.  (Indeed,
even if the Death Star is 500,000 kilometers from the planet when it
explodes, the planet's going to receive over 300,000 megatons of energy
per square kilometer that's exposed.)  Of course, the Empire is then
without a Death Star.  But if the planet it was attacking happened to be
the earth, the Federation's not going to be too happy either...

As for the quantum black holes...  Even if they were available,
channeling the beam back at the Death Star might not be such a hot idea
(see the previous paragraph).  A Death-Star-busting phaser would
similarly be a suicide weapon for the planet involved.  A better strategy
might be to simply drop a sufficiently-massive quantum black hole _into_
the Death Star (after all, it's not very maneuverable!).  I think this
might involve a radiation flux as the last of the Death Star's matter
was sucked in, but I lack the expertise to calculate it.

But suppose that, by whatever means, the Feds do take out the Death
Star.  What might be the Empire's next move?  Well, how about this... 
Instead of building a Death Star, why not simply build a robot bomb?  It
drops out of hyperspace near the target planet and immediately explodes! 
Even a blast of a few trillion megatons, at the distance the moon, would
utterly wipe out all life on one hemisphere of the planet (and those on
the other side had better get some life-support equipment pretty fast). 
Robot bombs being much easier to build than Death Stars, the Empire
could produce them relatively quickly.  So once again it looks like
curtains for the Federation unless the Q or somebody intervenes.

"Emperor, you just zorched my favorite human!  Just for that, I'm going
to turn all your Star Destroyers into Hershey bars!"

--
Leopold Travis
travis@alumni.caltech.edu
Newsgroups: rec.arts.sf.starwars